llMllllliliillilll 


H;iii!!niiii;iii 


llllii'llHIIMI 


a^3 


JN  \lEMORIAM 
FLORIAN  CAJORl 


^c      (E^^fL^^Tf^ 


Digitized  by  the  Internet  Archive 

in  2007  with  funding  from 

IVIicrosoft  Corporation 


http://www.archive.org/details/elementarygeometOOholgrich 


ELEMENTARY   GEOMETRY 
PLANE   AND   SOLID 


^s&^^ 


ELEMENTARY  GEOMETRY 

PLANE    AND    SOLID; 

FOR  USE  IN  HIGH  SCHOOLS  AND  ACADEMIES 


BY 


THOMAS    F.    HOLGATE 
♦i 

PROFESSOR  OF  APPLIED   MATHEMATICS   IN   NORTHWESTERN 
UNIVERSITY 


THE   MACMILLAN   COMPANY 

LONDON:  MACMILLAN  &  CO.,  Ltd. 
1901 

All  rights  reserved 


Copyright,  1901, 

bt  the  macmillan  company. 


CAJORl 

»    «  •        *  *    '   <  • 


Noriuoot)  i^rres 

J.  S.  Gushing  &  Co.  —  Berwick  &  Smith 

Norwood  Masa.  U.S.A. 


PREFACE 

Elementary  Geometry  deals  only  with  forms  whose  deter- 
mining parts  are  points,  straight  lines,  and  circles.  Its  method 
is  that  employed  by  the  ancient  writers.  The  amount  of  such 
study  to  be  included  in  the  high  school  or  academy  course  has 
been  fixed  by  tradition,  as,  in  plane  geometry,  the  equivalent 
of  Euclid's  first  six  books  together  with  some  additional 
material  on  mensuration ;  and,  in  solid  geometry,  the  equiva- 
lent of  Euclid's  eleventh  and  twelfth  books,  to  which  is  also 
added  the  mensuration  of  solids.  Modern  concepts  and  modern 
methods  have  given  to  this  old  material  a  correlation  and  a 
symmetry  which  it  did  not  at  one  time  present,  and  have 
opened  fields  of  investigation  which  were  entirely  beyond  the 
range  of  the  early  geometers.  From  the  modern  point  of  view, 
many  isolated  and  apparently  independent  theorems  have 
proved  to  be  but  special  cases  of  broader  and  more  general 
ones,  or  to  be  related  to  eaxjh  other  in  an  easily  defined  way. 
The  so-called  modern  geometry  possesses  great  beauty  and 
strength,  but  how  much  of  it  can  be  wisely  woven  into  a  first 
course  is  a  matter  about  which  there  is  no  general  concensus  of 
opinion.  Some  recent  writers  have  deemed  it  wise  to  introduce 
general  principles  very  early,  while  others  have  held  rigorously 
to  the  old  methods  and  old  materials.  My  own  belief  is  that 
the  pupil  must  become  quite  familiar  with  the  incidental  and 
particular  facts  of  geometry  before  he  is  capable  of  much 
generalization,  and  while  I  have  written  with  the  modern  notions 
distinctly  in  mind,  I  have  preferred  not  to  depart  far  from  the 
well-beaten  path. 

The  attempt  has  been  made  to  arrange  the  contents  of  this 


918229 


VI  PREFACE 

book  in  a  natural  and  attractive  pedagogical  order.  After  the 
introductory  chapter  there  are  presented  a  few  easy  problems 
of  construction  in  which  the  pupil  will  make  free  use  of  his 
ruler  and  compasses,  and  be  introduced  to  the  idea  of  a  formal 
proof  in  connection  with  matters  which  he  clearly  sees  need 
proving;  other  similar  problems  are  inserted  where  they  come 
in  most  naturally  and  are  of  most  immediate  interest. 

Abstract  discussions,  such  as  the  theory  of  limits,  have  been 
postponed  as  far  as  possible  in  order  to  secure  for  their  com- 
prehension greater  mathematical  maturity.  Young  pupils  do 
not  readily  assimilate  theoretical  or  abstract  principles,  and  it 
is  best  that  these  should  be  brought  into  but  little  prominence 
in  the  early  parts  of  the  subject. 

The  exercises  have  been  carefully  selected  and  are  intended 
to  form  an  integral  part  of  the  work.  They  bear  a  direct 
relation,  in  most  cases,  to  the  propositions  with  which  they 
are  associated ;  those  inserted  at  the  ends  of  sections  or  chap- 
ters are  for  the  most  part  of  a  more  difficult  character  and  may 
be  omitted  on  a  first  reading. 

It  is  hoped  that  the  classified  summary  at  the  end  of  each 
chapter  will  prove  serviceable  for  purposes  of  review  and  for 
reference,  and  will  also  aid  the  pupil  in  systematizing  his 
knowledge. 

Care  has  been  taken  to  state  the  fundamental  assumptions 

—  the  postulates  —  upon  which  the  science  rests,  in  as  clear 
a  form  as  possible,  and  to  distinguish  between  assumption  and 
axiomatic  truth.  Some  things  have  been  assumed  which  are 
often  made  matters  of  demonstration;  for  example,  the  fact 
that  the  perimeter  of  a  regular  polygon  inscribed  in  a  circle 
approaches  a  limit  as  the  number  of  its  sides  is  indefinitely 
increased.  This  admits  of  a  rigorous  proof,  to  be  sure,  but 
the  proofs  given  in  text-books  on  elementary  geometry  are 
as  a  rule  either  unsatisfactory  or  beyond  the  appreciation 
of  the  pupil.     I  have  preferred  openly  to  assume  the  property, 

—  an  assumption  at  which  the  pupil  does  not  hesitate. 


PREFACE  Vll 

A  circle  has  been  defined  as  a  particular  kind  of  line,  and  a 
polygon  as  a  figure  made  up  of  points  and  lines.  The  area  of 
such  a  figure  is  defined  to  be  the  surface  enclosed  by  it.  This 
accords,  I  believe,  with  the  best  usage,  though  perhaps  not 
with  the  common  usage.  Everywhere  outside  of  a  class  in 
elementary  geometry  a  circle  is  so  understood.  Neither  in  more 
advanced  mathematics  nor  in  everyday  life  is  it  thought  of  ao 
a  portion  of  a  plane,  in  accordance  with  the  common  definition, 
and  I  see  no  reason  why  the  pupil  should  be  obliged  to  change 
his  idea  of  such  a  figure  upon  entering  the  geometry  class,  and 
change  back  again  immediately  upon  leaving  that  class. 

The  Appendix  contains  a  short  chapter  on  Plane  Trigo- 
nometry, which  is  intended  to  serve  as  a  brief  introduction 
to  the  subject,  to  meet  the  needs  of  those  preparatory  pupils 
who  take  up  the  study  of  physics  or  mechanics,  not  as  a 
substitute  for  a  complete  course.  It  may  be  found  sufficient 
also  to  meet  the  practical  needs  of  those  who  do  not  continue 
their  studies  beyond  the  high  school. 

The  teacher  will  do  well  to  consider  the  following  sugges- 
tions :  — 

1.  Eead  the  introductory  chapter  carefully,  then  talk  it 
over  with  the  class  in  an  informal  way,  calling  attention  to 
the  geometrical  principles  involved,  most  of  which  will  be 
readily  accepted  by  your  pupils.  Do  not  assign  this  chapter 
as  a  lesson. 

2.  Proceed  very  slowly  at  first.  Eemember  that  your  pupils 
already  have  some  geometrical  ideas.  Draw  these  out,  clarify 
and  fix  them.  Do  not  break  down,  but  build  on  what  your 
pupils  already  have. 

3.  Do  not  be  too  strenuous  at  first  about  a  formal  demon- 
stration. Emphasize  the  geometric  truth  presented.  Fix  as 
your  ideal  an  elegant,  faultless  proof,  and  gradually  work 
up  to  it. 


Vlll  PREFACE 

4.  Eemember  that  in  this  subject  the  primary  object  should 
be  the  acquisition  of  geometric  knowledge  and  the  develop- 
ment of  the  geometric  sense.  Logical  reasoning  and  rhetorical 
demonstration  are  secondary  aims,  to  be  sure,  but  the  first 
object  should  be  Geometry. 

Since  my  chief  desire  has  been  to  produce  a  text-book 
adapted  to  the  needs  of  the  class-room,  I  have  not  hesitated 
to  make  free  use  of  many  existing  texts,  both  old  and  new, 
and  from  them  have  derived  much  valuable  help  and  many 
suggestions.  The  exercises  in  particular  have  been  gathered 
from  a  variety  of  sources;  only  a  few  of  them  are  new.  To 
many  friends,  and  in  particular  to  my  colleague.  Professor 
Henry  S.  White,  to  Mr.  B.  Annis,  of  the  Hartford,  Connecticut, 
High  School,  and  to  Mr.  J.  F.  Petrie,  of  North-western  Uni- 
versity Academy,  I  am  greatly  indebted  for  valuable  sugges- 
tions and  criticisms.  Also  to  my  pupil.  Miss  Elda  L.  Smith 
of  Springfield,  Illinois,  my  sincerest  thanks  are  due  for  patient 
and  careful  work  in  testing  all  exercises  and  in  the  tedious 
task  of  proof-reading. 

I  shall  be  glad  to  be  informed  of  any  errors  that  may  have 

been  overlooked,  and  to  receive  suggestions  for  improvement 

either  in  matter  or  arrangement. 

THOMAS  F.  HOLGATE. 

EvANSTON,  Illinois, 
May  30, 1901. 


CONTENTS 

PART   I 

PLANE    GEOMETRY 

INTRODUCTION 

PAGE 

Preliminary  Notions  and  Definitions 1 

Classified  Summary   . 19 

CHAPTER   I 

Triangles  and  Parallelograms 

Section     I.     General  Properties  of  Triangles 22 

Section    II.     Parallel  Lines 61 

Section  III.     Closed  Rectilinear  Figures.     Parallelograms        .         .  71 

Section  IV.     Miscellaneous  Theorems 78 

Section    V.     Additional  Problems  of  Construction   ....  84 

Section  VI.     Symmetrical  Figures 88 

Miscellaneous  Exercises 90 

Classified  Summary 92 


CHAPTER   11 

The  Circle 

Section     I.     Definitions  and  Preliminary  Theorems         ...  98 
Section    II.     Angles  inscribed  in  Arcs       .         .         .         .         .         .111 

Section  III.     Secants  and  Tangents.     Principle  of  Continuity  .  115 

Section  IV.     Circles  in  Contact 130 

Miscellaneous  Exercises 134 

Classified  Summary 136 

ix 


X  CONTENTS 

CHAPTER  III 
Similar  Rectilineak  Figures 

PAGE 

Section     1.     On  Measurement,  Ratio,  and  Proportion.     Theory  of 

Limits 140 

Section    II.     Similar  Polygons 156 

Section  III.     Problems  of  Construction     .         .         .        .         .         .  174 

Section  IV.     Additional  Propositions.     Harmonic  Division       .         .180 

Miscellaneous  Exercises 189 

Classified  Summary 190 

CHAPTER   IV 

Areas  of  Plane  Polygons 

Section     I.     Parallelograms  and  Triangles 194 

Section    II.     Areas  of  Similar  Polygons 207 

Section  III.     Problems  of  Construction 220 

Section  IV.     Maxima  and  Minima 226 

Miscellaneous  Exercises 229 

Classified  Summary 230 

CHAPTER   V 
Measurement  of  the  Circle 

Section     I.     Regular  Polygons 233 

Section    II.     Measurement  of  the  Circle   .         .         .        .         .         .  244 

Section  III.     Problems  of  Construction 251 

Miscellaneous  Exercises 258 

Classified  Summary 260 

PART   II 


SOLID    GEOMETRY 

CHAPTER  VI 

Lines  and  Planes  in  Space 

Section     I.     Intersecting  Planes.     Parallels  and  I'erpendiculars      .  263 

Section    II.     Dihedral  Angles 284 

Section  III.     Polyhedral  Angles 296 

Miscellaneous  Exercises     .        .        . 304 

Classified  Summary 306 


CONTENTS  XI 

CHAPTER  VII 
Prisms  and  Pyramids 

PAGE 

Section     I.     Area  and  Volume  of  a  Prism 310 

Section    II.     Pyramids 328 

Section  III.     Similar  Polyhedrons 339 

Section  IV.     Regular  Polyhedrons 342 

Section    V.     Polyhedrons  in  General 345 

Miscellaneous  Exercises 347 

Classified  Summary 349 

CHAPTER   VIII 

Cylinders  and  Cones 

Section     I.     Cylinders 353 

Section    II.     The  Cone 361 

Miscellaneous  Exercises 366 

Classified  Summary 368 

CHAPTER   IX 

The  Sphere 

Section     I.     Plane  Sections  and  Tangent  Planes      ....  370 

Section    II.     Spherical  Angles,  Triangles,  and  Polygons  .        .         .  380 

Section  III.     Polar  Triangles 387 

Section  IV.     Areas  of  Spherical  Triangles         .        .         .        .         .  394 

Section    V.     Volume  of  the  Sphere 405 

Miscellaneous  Exercises 409 

Classified  Summary 410 

APPENDIX 

Plane  Trigonometry        .........  416 

TABLES 

Mensuration  Formulas 435 

Index  to  Definitions 437 


PART   I— PLANE    GEOMETRY 

INTRODUCTION 
PRELIMINARY  NOTIONS  AND  DEFINITIONS 

When  we  speak  of  a  point  or  a  line,  a  straight  line  or  a 
crooked  line,  everybody  knows  in  a  general  way  what  is 
meant.  In  the  minds  of  persons  not  familiar  with  the  strict 
sense  in  which  these  terms  are  used  in  Geometry,  it  may  be 
that  the  words  *  point '  and  '  line '  are  associated  with  marks  of 
some  sort,  made  for  example  with  a  pen  or  pencil ;  they  may 
think  of  a  point  as  a  dot,  and  of  a  line  as  a  long,  narrow 
mark,  but  in  spite  of  this  association  the  true  meaning  of  the 
terms  'point,'  Mine,'  'straight,'  'crooked,'  etc.,  is  pretty  clearly 
understood. 

It  is  not  at  all  certain  that  definitions  could  be  given  in 
words  which  would  convey  an  intelligent  idea  of  the  meaning 
of  these  terms  to  a  person  not  already  familiar  with  the  thought 
conveyed  by  them,  still  a  few  words  of  comment  may  serve  to 
fix  the  attention  upon  some  properties  of  points  and  lines 
which  are  fundamental  in  the  study  of  Geometry. 

1.   A  point  has  no  magnitude ;  it  has  only  position. 

The  smallest  mark  that  can  be  made  with  the  finest  pencil 
or  pen  will  have  some  magnitude,  and  consequently  is  not  a 
true  point  in  the  sense  in  which  this  term  is  used  in  Geometry. 
We  can  thin\of  points  as  having  position  only,  but  in  making 
diagrams  we  are  obliged  to  use  pencil  or  pen  marks  to  repre- 
sent them. 

B  1 


ELEMENTARY  GEOMETRY 


[Introd. 


2.   A  line  has  position  and  length,  but  no  breadth  or  thickness. 

Whenever  we  speak  of  a  line  it  suggests  the  idea  of  exten- 
sion in  one  way,  namely,  length,  without  regard  to  breadth  or 
thickness;  but  if  we  try  to  represent  a  line  by  a  mark,  this 
mark  will  of  necessity  have  some  breadth  and  so  cannot  be  a 
true  geometrical  line.  You  should  endeavor  to  think  of  a  line 
without  actually  drawing  a  mark  to  represent  it. 

A  line  is  sometimes  spoken  of  as  the  path  of  a  moving  point. 

On  a  line  you  can  choose  as  many  points  as  you  please. 
Through  two  points  you  can  draw  as  many  lines,  of  one  kind 
or  another,  as  you  please. 


•D 


3.  In  order  to  tell  one  point  from  another  we  give  them 
names,  conveniently  the  names  of  the  letters  of  the  alphabet, 
and  place  the  letter  by  the  side  of  the  point  bear- 
ing its  name. 

Thus  we  speak  of  the  point  A,  the  point  B,  etc. 

If  we  wish  to  designate  different  points  by  the 
same  letter,  P  say,  we  distinguish  them  as  Pj,  P^y  Ps,  and 
so  on. 


•o 


•B 


We  denote  a  line  by  naming  a  sufficient  number  of  points  on 
it  to  distinguish  it  from  every  other  line. 

Thus  we  have  in  the  diagram  the 
line  ABC,  the  line  DE,  the  line  FBE. 

The  two  points  B  and  E  would  not 
distinguish  the  line  DBE  from  the  line 
FBE,  since  B  and  E  are  points  of  both 
lines,  or  as  we  say,  are  common  to  both 
lines.  The  lines  PQ  and  RS  in  the  second  diagram  are,  how- 
ever, distinguished  by  naming  two  points  on  each. 


2-5]  INTRODUCTION  3 

4.  A  straight  line  is  such  that  only  one  of  that  kind  can 
pass  through  the  same  two  points. 

That  is  to  say,  if  two  straight  lines  pass  through  the  same 
two  points,  they  must  occupy  the  same  position,  and  there 
is  no  point  of  one  which  does  not  also  lie  on  the  other. 

This  is  sometimes  expressed  by  saying  that,  "  If  two  straight 
lines  coincide  at  two  points,  they  coincide  throughout";  or, 
"  Two  points  determine  a  straight  line,"  i.e.  fix  the  location  of 
the  line  and  distinguish  it  from  every  other  straight  line. 

Two  different  straight  lines  therefore  can  have  only  one 
point  in  common ;  in  other  words,  two  straight  lines  intersect 
in  only  one  point. 

Any  two  straight  lines  can  be  made  to  coincide  at  every 
point  by  placing  them  so  that  they  coincide  at  two  points. 

5.  Suppose  that  on  a  given  straight  line  we  choose  two  points 
and  mark  them  by  the  letters  A  and  B.  These  will  serve  to 
distinguish  this  straight  line 

from  every  other,  since  no  ^ £ 

other  straight  line  can  pass 

through  both  of  these  points.     We  may  therefore  call  this  line 

the  straight  line  AB, 

That  portion  of  the  line  which  lies  between  the  points  A  and 
B  is  called  the  segment  AB  of  the  line.  If  other  points  are 
chosen  on  the  line, 

as  C,  D,  E,  etc.,  a o B D_ 

the  portion  of  the 

line  lying  between  the  points  A  and  G  is  called  the  segment 

AC ;  that  between  C  and  D,  the  segment  CD^  etc. 

It  should  be  observed  that  two  points  chosen  anytbhere  on  a 
straight  line  determine  it;  the  line  may  be  as  long  as  you 
please,  i.e.  it  may  be  unlimited  in  extent,  yet  these  two  points 
will  distinguish  it  from  every  other  straight  line.  When  we 
have  in  mind  only  that  segment  of  a  straight  line  lying  between 


4  ELEMENTARY  GEOMETRY  [Introd. 

two  fixed  points  A  and  B,  we  shall  usually  speak  of  it  as  the 
line-segment  AB. 

The  distinction  thus  made  between  tJie  straight  line  and  the 
line-segment  is  sometimes  effected  by  calling  the  former  an 
unlimited  straight  line,  and  the  latter  a  terminated  straight  line. 
Of  course  any  mark  that  we  make  to  represent  a  straight 
line  must  be  terminated  both  ways  by  points;  but  the  term 
'  line-segment '  will  be  used  only  when  we  wish  to  restrict  our 
thought  to  that  portion  of  the  whole  line  which  lies  between 
two  particular  points. 

6.  When  two  line-segments  having  one  end-point  in  com- 
mon lie  in  the  same  straight 

line  and  do  not  overlap,  for       -A C_ £ 

example,  the  segments  AG  and 

CB,  then  the  segment  which  is  made  up  of  these  two,  i.e.  the 

segment  AB,  is  called  their  sum. 

If  the  line-segments  have  one  end-point  in  common  and  do 
overlap,  then  that  segment  of  the  one  which  is  not  covered  by 
the  other  is  called  their  difference. 

Thus  the  difference  of  the  segments  AB  and  GB,  in  the  last 
diagram,  is  the  segment  AG. 

7.  A  surface  extends  in  two  ways,  i.e.  it  has  length  and 
breadth,  but  no  thickness. 

We  speak  of  the  surface  of  the  blackboard,  the  surface  of 
the  table,  the  surface  of  the  earth,  and  think  of  their  extension 
in  two  ways  only,  not  at  all  of  depth  or  thickness.  A  geo- 
metrical surface  has  no  thickness. 

Any  mimber  of  lines  can  be  drawn  on  a  surface,  and  any 
number  of  points  can  be  chosen  on  a  surface. 

If  a  surface  is  such  that  every  straight  line  joining  two 
points  of  it  lies  wholly  on  the  surface,  it  is  called  a  plane 
surface. 


5-10]  INTRODUCTION  5 

The  top  of  a  table  is  approximately  a  plane  surface,  while 
the  surface  of  an  apple  is  clearly  not  plane. 

For  brevity  we  sometimes  use  the  term  a  plane  instead  of 
a  plane  surface. 

8.  A  solid  extends  in  three  ways,  i.e.  it  has  length,  breadth, 
and  thickness. 

9.  A  geometrical  figure  is  any  combination  of  points,  lines, 
and  surfaces. 

A  plane  geometrical  figure  is  one  whose  points  and  lines  all 
lie  in  one  plane. 

Triangles  and  Angles 

10.  One  of  the  simplest  geometrical  figures  is  what  is 
called  a  triangle.  It  consists  of  three  straight  lines  which 
intersect  two  and  two  in  three  points. 

The  three  points  are  called  the  vertices  of  the  triangle,  and 
the  three  straight  lines  its  sides. 


In  most  cases  when  we  speak  of  a  triangle  we  shall  have  in 
mind  only  the  figure  consisting  of  the  three  vertices  and  the 
three  line-segments  lying  between  them,  and  then  the  three 
line-segments  are  thought  of  as  the  sides  of  the  triangle.  It 
should  be  remembered,  however,  that  these  sides  may  be 
extended  as  far  as  we  wish  at  any  time, 


6 


ELEMENTARY  GEOMETRY 


[Introd. 


The  length  of  a  side  will  always  mean  the  length  of  the  line- 
segment  between  the  vertices. 

A  triangle  may  be  designated  by 
naming  its  vertices  since  no  differ- 
ent triangle  can  have  the  same  three 
vertices. 

Thus  the  triangle  in  the  diagram 
is  the  triangle  ABC,  or  the  triangle 
BAC. 

Exercise.     Name  the  triangle  in  the  diagram  in  six  different  ways. 

Sometimes  it  is  found  convenient  to 
designate  a  straight  line  by  a  single  let- 
ter, and  in  this  case  we  commonly  use  a 
small  letter,  reserving  the  capital  letters 
to  denote  points.  When  the  sides'  of  a 
triangle  are  so  marked  we  can  designate 
the  triangle  by  naming  its  three  sides. 


11.  In  looking  at  the  figure  of  a  triangle  it  will  be  noticed 
that  the  two  sides  which  meet  in  any  vertex  start  out  from  the 
vertex  in  different  directions. 

Whenever  tw^o  straight  lines  which  meet  diverge  from  their 
common  point  in  this  way,  they  are  said  to  form  a  plane  angle, 
or  simply  an  angle,  with  each  other. 

The  idea  of  an  angle  will  be  made  clearer  by  the  accompany- 
ing figure.  The  straight  lines  AB  and 
AC  meet  at  the  point  A.  They  make 
with  each  other  an  angle  at  this  point. 
This  angle  we  name  the  angle  BAC,  or 
the  angle  CAB,  meaning  by  that  the  angle 
formed  by  the  lines  BA  and  AC  It  is 
sometimes  spoken  of  as  "  the  angle  between  the  lines  AB  and 
u4(7,"  or  "the  angle  contained  by  the  lines  AB  and  J.O." 

If  we  keep  the  line  AB  fixed  in  position  and  rotate  the  line 


10-12]  INTRODUCTION  7 

AC  about  the  point  A  in  the  way  indicated  by  the  arrowhead, 
we  enlarge  the  angle  BAG]  that  is,  we  increase  the  divergence 
between  the  lines  AB  and  AC,  and  so  enlarge  the  angle.  If, 
however,  we  rotate  the  line  the  other  way,  the  divergence  be- 
comes less;  that  is,  the  angle  formed  by  the  lines  becomes 
smaller. 

The  size  of  the  angle  formed  by  two  lines  does  not  depend  on 
the  length  of  the  lines,  or  on  anything  except  the  amount  of 
their  divergence. 

The  straight  lines  AB  and  AG  drawn  from  A  are  called  the 
boundaries  of  the  angle,  and  their  point  of  intersection  A,  the 
vertex  of  the  angle. 

It  will  readily  be  seen  that  there  are  three  angles  in  any 
triangle,  namely,  the  angle  made  by  the  sides  AB  and  AG  at 
A,  that  made  by  the  sides  BA  and  BG  at  B,  and  that  made  by 
the  sides  GA  and  GB  at  G.  For  brevity  we  shall  sometimes 
speak  of  these  as  the  angle  A,  or  the  angle  B,  or  the  angle 
G,  but  it  must  always  be  borne  in  mind  that  we  mean  the  angle 
at  A,  or  B,  or  O,  made  by  the  two  straight  lines  which  meet  at 
that  point. 

The  pupil  should  carefully  observe  that  in  reading  an  angle 
the  letter  at  the  vertex  is  always  named  between  the  other  two 
letters.  The  angle  BAG  and  the  angle  GAB  are  the  same, 
having  the  same  vertex  at  A  and  the  same  boundaries  AB  and 
AG,  while  the  angle  ABG  is  different,  having  its  vertex  at  B. 


12.  When  two  angles  are  placed  so  as  to  have  the  same  point 
for  vertex  and  one  boundary  in  common  without  overlapping, 
the  angle  formed  by  the  other  two  boundaries,  of  which  these 
two  angles  are  parts,  is  called  the  sum  of  the  two  angles. 

If,  on  the  other  hand,  the  two  angles  have  the  vertex  and 
one  boundary  in  common  and  do  overlap,  then  the  angle  formed 
by  the  other  two  boundaries  is  called  the  difference  of  the  two 
given  angles. 


8 


ELEMENTARY  GEOMETRY 


[Introd. 


For  instance,  the  angle  BAD  in  the  figure  is  the  sum  of  the 
angle  ^^(7 and  the  angle  GAD, 
while  the  angle   BAC  is  the 
difference  of  the  angle  BAD 
and  the  angle  CAD. 

When  two  angles  have  a  com- 
mon vertex  and  one  common 
boundary,  without  overlap- 
ping, they  are  called  adjacent 
angles. 

Thus  in  the  figure  the  angle  BAC  and  the  angle  CAD  are 
adjacent  angles. 

13.  Suppose  that  on  a  straight  line  AB  we  choose  any  point  (7, 
and  from  it  draw  another  straight  line  CD  (Fig.  1).  This  line 
makes  two  adjacent  angles 

with  AB,  namely,  the  angle  /D 

BCD  and  the  angle  DC  A. 
If  the  line  be  drawn,  as  ED 
(Fig.  2),  so  as  to  cross  AB, 
then  it  also  makes  with  AB 
two  adjacent  angles  on  the 
other  side,  namely,  the  angle 
ECB  and  the  angle  ECA. 

In  Fig.  2  the  angles  whose 
boundaries  lie  in  the  same 
straight  line,  but  which  have 
only  their  vertices  in  com- 
mon, for  example,  the  angle 
ACE  and  the  angle  BCD, 
are  called  vertically  opposite 
angles,  or  briefly,  vertical 
angles. 

Exercise.  In  this  figure  point  out  and  name  all  pairs  of  adjacent 
angles  and  all  pairs  of  vertically  opposite  angles.  How  many  pairs  of 
each  kind  do  you  find  ? 


C 
Fig.  1 


Fig.  2 


12-15]  INTRODUCTION  9 

Test  of  Equality 

14.  In  Elementary  Geometry  we  assume  that  figures  can  be 
moved  about  in  any  manner  without  altering  their  size  or 
shape.  Upon  this  assumption  we  have  the  following  tests  of 
equality : 

1.  If  two  line-segments  can  be  so.  placed  that  their  end- 
points  coincide,  they  are  equal;  and  conversely,  if  two  line- 
segments  are  equal,  they  can  be  placed  so  that  their  end-points 
will  coincide. 

2.  If  two  plane  angles  can  be  so  placed  that  their  boundaries 
coincide,  they  are  equal ;  and  conversely,  if  two  plane  angles 
are  equal,  they  can  be  placed  so  that  their  boundaries  coincide. 

3.  If  two  triangles  can  be  so  placed  that  their  vertices  coin- 
cide, and  consequently  also  their  sides  and  angles,  they  are 
identically  equal ;  and  conversely,  if  two  triangles  are  identi- 
cally equal,  they  can  be  placed  so  that  their  vertices,  their 
sides,  and  their  angles  will  coincide. 

In  general,  two  geometrical  figures  are  said  to  be  identically 
equal  when,  and  only  when,  they  can  be  made  to  occupy,  point 
for  point,  and  line  for  line,  the  same  positions. 


Right  Angles  and  Straight  Angles 

15.    Suppose  that  from  any  point  (7  in  a  straight  line  AB,  a 
straight  line  CD  is  drawn  so  as  to  make 
the  adjacent  angles  BCD  and  ACD  equal,  ^ 

then  each  of  these  equal  angles  is  called 
a  right  angle,  and  CD  is  said  to  be  at 

right  angles  to  AB,  or  perpendicular  to      

AB. 


Definition.  If  one  straight  line  meets  another  so  as  to 
make  the  two  adjacent  angles  equal,  each  of  these  angles  is 
called  a  right  angle. 


10  ELEMENTAHY  GEOMETRY  [Introd. 

16.  Let  us  take  any  straight  line  AB  and  another  straight 
line  CD  meeting  it  at  the  point  C  so  as  to  make  two  adja- 
cent angles  ACD  and  BCD.  If 
now  we  let  the  line  AB  remain  ^ 

fixed  in  position,  and  rotate  the  ^  \ 

line    CD  about   the   point    C  so  ^^ 

that  it  takes  up  successively  the     

positions  CD^,  CD^,  CD^,  etc.,  the  ^  "" 

effect  will  be  to  gradually  increase  one  of  the  two  adjacent 
angles  made  by  the  lines,  and  to  decrease  the  other. 

The  sum  of  the  two  angles  however  will  remain  unaltered, 
since  the  boundaries  of  the  sum,  viz.  CB  and  CA,  are  not 
changed  by  the  rotation  of  CD. 

In  some  one  position,  as  at  CD^,  each  of  the  adjacent  angles 
is  aright  angle,  and  their  sum  is,  therefore,  two  right  angles. 
Hence  in  whatever  position  the  line  CD  is  drawn  the  sum  of 
the  two  adjacent  angles  BCD  and  ACD  is  always  equal  to 
two  right  angles. 

If  the  line  CD  should  cross  AB,  the  sum  of  the  two  angles 
on  the  other  side  of  AB  would  also  be  equal  to  two  right  angles. 

That  is  to  say, 

TJie  sum  of  the  tivo  migles  ivhich  one  straight  line  makes  with 
another  on  one  side  of  it  is  equal  to  two  right  angles;  and  when 
tivo  straight  lines  intersect,  the  sum  of  the  four  angles  formed  at 
their  point  of  intersection  is  equal  to  four  right  angles. 


17.    The  pupil  should  notice  carefully  the  argument  which  led  up  to 
the  above  conclusion. 

(1)  The  straight  line  CD  makes  two  adjacent  angles  with  the  straight 
line  AB.,  which  vary  in  size  as  CD  rotates  about  C. 

(2)  The  sum  of  these  adjacent  angles  does  not  vary  by  the  rotation 
of  GD. 

(3)  For  one  position  of  CD,  each  of  the  adjacent  angles  is  a  right 
angle,  and  consequently  their  sum  is  equal  to  two  right  angles. 

(4)  Therefore  for  every  position  of  CD,  the  sum  of  the  two  adjacent 
angles  is  equal  to  two  right  angles. 


16-20]  INTRODUCTION  11 

Finally,  the  argument  which  applies  to  the  two  angles  on  one  side  of 
AB  would  apply  equally  well  to  the  two  angles  on  the  other  side,  if  the 
line  CD  were  drawn  so  as  to  cross  AB  at  C. 

18.  From  Article  16  it  follows  at  once  that  —  If  from  a 
point  in  a  straight  line  any  number  of  straight  lines  are  drawn 
in  a  plane  on  one  side  of  the  line,  the  sum  of  all  the  angles  so 
formed  is  equal  to  two  right  angles;  and, 

If  any  number  of  straight  lines  are  draivn  from  a  point  in  a 
plane,  the  sum  of  all  the  angles  so  formed  is  equal  to  four  right 
angles. 

19.  When  the  sum  of  two  angles  is  equal  to  two  right 
angles,  each  of  them  is  called  the  supplement  of  the  other. 

If  two  supplementary  angles  are  not  equal,  one  of  them  must 
be  less  than  a  right  angle,  and  the  other,  greater  than  a  right 
angle. 

An  angle  which  is  less  than  a  right  angle  is 
called  an  acute  angle. 

An  angle  which  is  greater  than  a  right  angle 
and  less  than  two  right  angles  is  called  an  obtuse     \ 
angle.  \  ^^ 

An  acute  angle  is  always  less  than  its  sup- 
plement, while  an  obtuse  angle  is  greater  than  its  supplement. 

The  supplement  of  an  acute  angle  is  obtuse,  and  the  supple- 
ment of  an  obtuse  angle  is  acute. 

What  is  the  supplement  of  a  right  angle  ? 

20.  Suppose  again  we  let  one  boundary  OB  of  an  angle 
BOG  remain  fixed  while  we 

rotate    the    other    boundary 

OC  about  the  vertex  so  as  to  \ 

increase  the  angle.     In  one  \^ 

position,    say    the    position  ^ 

OCi,  the  two  boundaries  are  '  ^,-'''''  ^ 

in  the  same  straight  line. 


c 


12  ELEMENTARY  GEOMETRY  [Introd. 

The  angle  BOC  has  then  become  a  so-called  straight  angle, 
and  if  OC  continues  to  rotate  beyond  this  position  the  angle 
becoiues  reflex. 

Definitions.  A  straight  angle  is  an  angle  whose  bound- 
aries extend  from  the  vertex  in  opposite  ways,  and  lie  in  the 
same  straight  line. 

A   reflex   angle   is   an  ,''"■"-. 

angle   which    is    greater  i — .— i 

than  a  straight  angle. 

A    straight    angle     is  ,'— -^ 

equal  to  the  sum  of  two 
right  angles ;  in  other 
words,  a  right  angle  is 
half  of  a  straight  angle. 

21.  If  we  place  one  straight  angle  upon  another  so  that 
their  vertices  coincide,  and  so  that  one  boundary  of  the  first 
shall  coincide  with  one  boundary  of  the  second,  then  the  other 
boundary  of  the  first  must  also  coincide  with  the  other  bound- 
ary of  the  second.  For  the  two  boundaries  of  each  angle  lie  in 
a  straight  line,  and  these  straight  lines  have  been  made  to 
coincide  in  more  than  one  point.  Hence  they  must  coincide 
throughout.     Art.  4. 

Therefore  any  one  straight  ayigle  is  equal  to  any  other  straight 
angle;  or,  all  straight  angles  are  equal. 

Since  a  right  angle  is  always  half  of  a  straight  angle,  and  all 
straight  angles  are  equal,  therefore  all  right  angles  are  equal. 

The  sum  of  any  angle  and  its  supplement  is  a  straight  angle. 
Hence,  if  two  angles  are  equal  their  supplements  are  equal,  and 
two  different  supplements  of  the  same  angle  are  equal. 

22.  Let  us  now  take  two  supplementary  angles  AOC  and 
BOC  and  place  them  so  that  they  will  be  adjacent,  that  is,  so 
that  their  vertices  will  coincide  and  one  boundary  of  one  will 
fall  on  one  boundary  of  the  other,  while  the  angles  do  not 
overlap. 


20-23] 


INTRODUCTION 


13 


The  other  boundaries  AO  and  OB  must  then  be  parts  of  the 
same  straight  line. 

For,  if  we  should  prolong  '^ 

the  straight  line  AO  through 
the  vertex,  the  angle  so 
formed  with    CO   would  be 

the  supplement  of  ^OC.    But     ^ ' 

BOC  is  the  supplement  of 

AOC.     Therefore  OB  must  coincide  with  the  extension  oi  AO 

through  the  vertex.     That  is  to  say, 

//'  ti(jo  adjacent  angles  are  supplementary ^  their  outer  boundaries 
lie  in  the  same  straight  line. 


On  Closed  Lines 

23.  By  a  closed  line  we  mean  one  that  can  be  followed  with 
a  pencil  or  a  pen,  starting  from  any  point  of  it 
and  returning  to  the  same  point,  without  remov- 
ing the  pencil  or  pen  from  the  paper,  and  with- 
out passing  twice  over  the  same  portion  of  the 
line. 

Any  figure  which  can  be  traced  in  a  similar 
way,  starting  at  one  point  and  returning  to  the  same  point,  is 
called  a  closed  figure. 


If  a  closed  figure  is  made  up  wholly  of  segments  of  straight 
lines,  it  is  called  a  closed  rectilinear  figure. 

Any  connected  set  of  line- 
segments  which  is  not  closed 
may  be  called  a  broken  line. 


\ 


14  ELEMENTARY  GEOMETRY  [Introd. 


Properties  of  Closed  Lines 

24.  If  a  point  is  chosen  inside  of  any  closed  line,  every 
straight  line  drawn  through  it  must,  if  extended  far  enough, 
cut  the  closed  line  in  at  least  two  points. 

If  one  closed  line  lies  partly  inside  and  partly  outside  of 
another  closed  line,  these  two  closed  lines  must  intersect  in  at 
least  two  points. 

25.  Definition.  A  circle  is  a  closed  line  so  drawn  that 
all  points  of  it  are  equally  distant  from  a  certain  fixed  point 
within  it. 

This  fixed  point  is  called  the  centre  of  the  circle. 

A  line-segment  drawn  from  the  centre  to  any  point  of  the 
circle  is  called  a  radius  of  the  circle,  while  a  line-segment  drawn 
through  the  centre  and  terminated  both  ways  by  the  circle  is 
called  a  diameter  of  the  circle. 

A  diameter  of  a  circle  is  twice  as  long  as  a  radius,  or  is  equal 
to  the  sum  of  two  radii. 

If  a  straight  line  intersects  a  circle,  it  must  cut  it  at  two 
points  if  extended  far  enough,  and  when  one  circle  lies  partly 
within  and  partly  without  another,  they  must  intersect  at  two 
points. 

On  the  use  of  Instruments 

26.  In  the  construction  of  figures  in  Elementary  Geometry 
it  is  customary  to  limit  ourselves  to  the  use  of  a  straight-edge 
or  ruler  and  a  pair  of  compasses. 

With  the  ruler  we  are  able  to  draw  a  straight  line  between 
any  two  points  and  also  to  extend  or  produce,  as  far  as  we 
please,  any  terminated  or  finite  straight  line  or  line-segment 
already  drawn. 

Tlie  ruler  is  supposed  to  have  no  scale  of  measurements 
marked  on  it,  so  that  it  cannot  be  used  to  compare  lengths. 


24-27]  INTRODUCTION  16 

The  compasses  are  used  for  describing  circles  and  for  mark- 
ing off  equal  lengths. 

The  pupil  will  readily  see  how  this  instrument  can  be  made 
use  of  — 

(1)  To  describe  a  circle  having  a  given  point  for  centre  and 
its  radius  equal  to  a  given  line-segment. 

(2)  To  mark  otf  on  the  longer  of  two  given  line-segments  a 
part  equal  to  the  shorter. 

(3)  To  draw  from  a  given  point  a  line-segment  equal  to  a 
given  line-segment. 

Postulates 

27.  In  geometry,  whenever  we  assume  a  fundamental  prop- 
erty or  the  possibility  of  constructing  some  simple  figure,  we 
say  that  we  postulate  these  things. 

A  few  of  these  fundamental  assumptions  or  postulates  have 
already  crept  into  our  discussions  along  with  the  definitions 
and  descriptions  of  simple  figures. 

Examples  op  Postulates 

(1)  A  straight  line  can  be  drawn  from  any  one  point  to  any 
other  point. 

From  the  defining  property  of  a  straight  line  there  is  only  one  such 
line.     Art.  4. 

(2)  A  terminated  straight  line,  or  a  line-segment,  can  be 
produced  to  any  length  in  the  same  straight  line. 

(3)  A  circle  can  be  described  having  any  given  point  for  centre 
and  its  radius  of  any  given  length. 

(4)  Geometrical  figures  can  be  moved  about  in  space  without 
altering  their  size  or  shape. 

Other  postulates  will  be  added  as  we  proceed,  and  these  will 
be  collected  in  tabulated  form  at  the  end  of  the  chapter  in 
which  they  first  appear. 


16  ELEMENTARY  GEOMETRY  [Introd. 

Axioms 

28.  A  simple  statement  or  proposition  which  is  readily 
admitted  to  be  true  as  soon  as  its  meaning  is  understood,  is 
called  an  axiom.  In  other  words,  an  axiom  is  a  statement  the 
truth  of  which  does  not  need  to  be  demonstrated,  and,  in  gen- 
eral, cannot  be  demonstrated ;  it  is,  as  we  say,  self-evident. 

Many  of  the  axioms  to  which  we  shall  make  reference  apply 
not  only  to  geometrical  magnitudes,  but  to  magnitudes  of  all 
kinds. 

The  term  '  magnitude '  is  applied  to  anything  that  has  size,  i.e.  to  any- 
thing that  can  be  increased  or  diminished.  For  example,  a  line-segment, 
an  angle,  the  surface  within  a  closed  figure,  are  magnitudes.  A  point  is 
not  a  magnitude,  since  it  has  no  size. 

The  following  are  a  few  examples  of  such  axioms,  but  it 
must  not  be  supposed  that  this  list  includes  all  that  could  be 
stated.  Other  axioms  will  be  announced  when  there  is  occa- 
sion for  their  use. 

Examples  of  Axioms 

1.  Magyiitudes  which  are  equal  to  the  same  magiiitude  are  equal 
to  each  other. 

If  A  equals  C,  and  B  equals  C,  then  A  equals  B. 

2.  If  equal  maynitudes  are  added  to  equal  magnitudes  the  sums 
are  equal. 

If  A  equals  B,  and  C  equals  Z>,  then  the  sum  of  A  and  C  equals  the 
sum  of  B  and  D. 

3.  If  equal  magnitudes  he  taken  from  equals,  the  remainders 
are  equal. 

4.  If  equal  magnitudes  he  added  to  unequals,  or  unequals  to 
equals,  the  sums  are  unequal. 

5.  If  equal  magnitudes  he  taken  from  unequals,  or  uyiequals 
from  equals,  the  remainders  are  unequal. 

6.  Doubles  of  the  same  magnitude,  or  of  equal  marinitudes,  are 
equal;  and  douhles  of  unequals  are  uneqiial. 


28-31]  INTRODUCTION  17 

7.  Halves  of  the  same  magnitude,  or  of  equal  magnitudes,  are 
equal;  and  halves  of  unequals  are  unequal. 

8.  The  whole  of  anything  is  greater  than  a  part  of  the  same 
thing. 

9.  If  one  magyiitude  is  greater  than  a  second,  and  the  second 
greater  than  a  third,  then  the  first  is  greater  than  the  third. 

29.  A  theorem  is  a  statement  of  fact  which  is  usually  not 
self-evident,  but  which  needs  to  be  demonstrated  before  its 
truth  becomes  apparent. 

For  example, 

(1)  If  two  adjacent  angles  are  supplementary,  their  outer  boundaries 
lie  in  the  same  straight  line.     Art.  22. 

(2)  The  sum  of  any  two  angles  of  a  triangle  is  less  than  two  right 
angles. 

A  theorem,  the  truth  of  which  is  self-evident,  has  already 
been  called  an  axiom. 

30.  To  demonstrate  a  theorem  is  to  establish  its  truth  by  a 
process  of  reasoning  in  which  reference  is  made  only  to  the 
accepted  definitions  and  axioms,  and  to  the  facts  contained  in 
previously  established  theorems. 

When  the  truth  of  a  theorem  has  once  been  established,  the 
theorem  is  used  without  further  question  whenever  it  will  be 
serviceable  in  the  demonstration  of  other  theorems. 

31.  A  problem  is  a  task  to  be  accomplished  by  the  use  of  the 
ruler  and  compasses. 

Tor  example, 

(1)  To  draw  a  straight  line  which  will  divide  a  given  angle  into  two 
equal  parts. 

(2)  To  construct  a  triangle  whose  sides  shall  be'equal  to  three  given 
line-segments. 

The  general  term  Proposition  is  used  to  include  both  theorems 
and  problems. 
c 


18  ELEMENTARY  GEOMETRY  [Introd. 

EXERCISES 

1.  Using  only  the  ruler  and  compasses,  draw  a  line-segment  which  shall 
be  (1)  twice  as  long  as  a  given  line-segment,  (2)  three  times  as  long  as  a 
given  line-segment. 

2.  Draw  a  line-segment  equal  in  length  (1)  to  the  sum  of  two  given 
line-segments,  (2)  to  the  difference  of  two  given  line-segments. 

3.  Describe  a  circle  whose  radius  is  (1)  equal  to  a  given  line-segment, 
(2)  double  of  a  given  line-segment. 

4.  Describe  two  circles  with  the  same  centre  such  that  the  radius  of 
one  is  equal  to  the  diameter  of  the  other. 

Definition.  Circles  having  the  same  centre  and  unequal  radii  are 
called  concentric  circles. 

5.  Describe  two  circles  so  situated  that  a  radius  of  one  is  a  diameter 
of  the  other. 

6.  Describe  two  circles  so  that  the  same  line-segment  is  a  radius  of 
each,  but  so  situated  that  the  circles  do  not  coincide. 

7.  Express  the  supplement  of  (1)  half  a  right  angle,  (2)  two-thirds  of 
a  right  angle,  in  terms  of  a  right  angle. 

8.  Draw  three  straight  lines  in  a  plane  so  that  they  do  not  all  pass 
through  one  point  and  each  intersects  the  other  two.  How  many  points 
of  intersection  are  there  ? 

9.  Choose  three  points  in  a  plane,  not  in  the  same  straight  line,  and 
join  each  point  to  both  the  others.     How  many  straight  lines  are  there  ? 

10.  Draw  four  straight  lines  in  a  plane  so  that  no  three  pass  through 
the  same  point  and  each  intersects  all  the  others.  How  many  points  of 
intersection  are  there  ? 

Name  the  lines  a,  &,  c,  <?,  and  mark  the  points  of  intersection  with 
capital  letters,  J[,  B^  C,  etc. 

Point  out  and  name  the  vertices  of  the  triangles  whose  sides  are  any 
three  of  the  four  lines.     How  many  such  triangles  are  there  ? 

To  how  many  of  these  triangles  does  each  line  belong  ? 

To  how  many  of  the  triangles  does  each  vertex  belong  ? 

11.  Choose  four  points  in  a  plane  so  that  no  three  lie  in  the  same 
straight  line,  and  join  each  point  to'  all  the  others.  How  many  lines  are 
thus  drawn  ?    Name  the  points  A,  B^  O,  Z>,  and  the  lines  a,  6,  c,  etc. 

Point  out  and  name  the  sides  of  all  the  triangles  whose  vertices  art 
any  three  of  the  four  points.     How  many  such  triangles  are  there? 
To  how  many  of  these  triangles  does  each  vertex  belong  ? 
To  how  many  of  these  triangles  does  each  line  belong  ? 


31]  INTRODUCTION  19 

SUMMARY 

1.  Simple  Elements. 

Points,  Lines,  Surfaces. 

2.  Properties  of  the  Simple  Elements. 

(1)  Point  —  has  no  magnitude  or  extension.     §  1. 

has  position  pnly.     §  1. 

(2)  Line  —  has  extension  in  one  way,  viz.  length.     §  2. 

path  of  a  moving  point.     §  2. 

contains  an  unlimited  number  of  points.     §  2. 

(3)  Surface  —  has  extension  in  two  ways,  viz.  length  and  breadth.  §  7. 

contains  an  unlimited  number  of  points  and  lines.  §  7. 

3.  Definitions. 

(1)  Straight  Line  —  aline  such  that  only  one  of  the  kind  can  pass 

through  two  points.     §  4. 

(2)  Plane  Surface  or  Plane  —  a  surface  such  that  every  straight  line 

joining  two  points  of  it  lies  wholly  on  the  surface.     §  7. 

(3)  Geometrical  Figure  —  any  combination  of  points,  lines,  and  sur- 

faces.    §  9. 

(4)  Equal  Geometrical  Figures  —  two  figures  which  can  be  made  to 

occupy  identically  the  same  position.     §  14. 

(5)  Line-segment  —  the  portion  of  a  straight  line  which  lies  between 

two  fixed  points.     §  5. 
Sum  or  Difference  of  Two  Line-segments  —  see  §  6. 

(6)  Angle  —  the  inclination  of  two  straight  lines  which  meet.     §  11. 

Boundaries  of  the  Angle  —  the  two  straight  lines  forming  it. 
Vertex  of  the  Angle — their  common  point. 
Sum  or  Difference  of  Two  Angles  —  see  §  12. 

(7)  Adjacent  Angles  —  two  angles  which  have  a  common  vertex  and 

one  common  boundary,  without  overlapping.     §  12. 

(8)  Vertically  Opposite  Angles  or  Vertical  Angles  —  two  angles  which 

have  a  common  vertex,  the  boundaries  of  each  being  the 
extensions  of  the  boundaries  of  the  other.     §  13. 

(9)  Bight  Angle  —  an  angle  formed  by  one  straight  line  meeting 

another  when  the  two  adjacent  angles  formed  by  them 
are  equal.     §  15. 
Acute  Angle,  Obtuse  Angle,  Reflex  Angle  —  see  §§19  and  20. 

(10)  Supplementary  Angles  —  two  angles  whose  sum  is  equal  to  two 

right  angles.     §  19. 

(11)  Straight  Angle  —  an  angle  whose  boundaries  extend  in  opposite 

ways  from  the  vertex  and  lie  in  the  same  straight  line.    §  20. 


20  ELEMENTARY  GEOMETRY  [Introd. 

(12)  Perpendicular  Lines  —  lines  which  form  a  right  angle.     §  15. 

(13)  Triangle  —  a  geometrical  figure  consisting  of  three  straight  lines 

intersecting,  two  and  two,  in  three  points.     §  10. 

Vertices  of  a  Triangle  —  the  points  in  which  the  three  lines 
intersect.     §  10. 

Sides  of  a  Triangle  —  the  three  lines,  the  line-segments  be- 
tween the  vertices.     §  10. 

Angles  of  a  Triangle  —  the  angles  formed  by  the  sides,  two 
and  two.     §  11. 

(14)  Circle  —  a  closed  line  so  drawn  that  all  points  of  it  are  equally 

distant  from  a  certain  fixed  point  within  it.     §  25. 
Centre  of  the  Circle  —  the  point  from  which  all  of  its  points 

are  equally  distant.     §  25. 
Radius  of  a  Circle  —  a  line-segment  drawn  from  the  centre  to 

any  point  of  the  circle.     §  25. 
Diameter  of  a  Circle  —  a  line-segment  drawn   through   the 

centre  and  terminated  both  ways  by  the  circle.     §  25. 

(15)  Concentric  Circles  —  those  having  the  same  centre  and  unequal 

radii.     Ex.  4,  p.  18. 

(16)  Postulate  —  a  fundamental  geometrical  property,  or  construction, 

which  is  assumed.     §  27. 

(17)  Axiom  —  a  statement  the  truth  of  which  is  admitted  as  soon  as 

its  meaning  is  understood.     §  28. 

(18)  Theorem  —  a  statement  of  fact  which  is  usually  not  self-evident, 

but  which  needs  to  be  demonstrated  before  its  truth  be- 
comes apparent.     §  29. 

(19)  Problem  —  a  task  to  be  accomplished  by  the  use  of  the  ruler  and 

compasses.     §  31. 

4.  Postulates. 

See  §  27  for  Postulates  1-4. 

5.  Axioms. 

See  §  28  for  Axioms  1-9. 

6.  Theorems. 

(1)  The  sum  of  the  two  angles  which  one  straight  line  makes  with 
another  on  one  side  of  it  is  equal  to  two  right  angles ;  and 
when  two  straight  lines  intersect,  the  sum  of  the  four  angles 
formed  at  their  point  of  intersection  is  equal  to  four  right 
angles.    §  16. 


Summary]  INTRODUCTION  21 

(2)  If  from  a  point  in  a  straight  line  any  number  of  straight  lines 

are  drawn  in  a  plane  on  one  side  of  the  line,  the  sum  of  all 
the  angles  so  formed  is  equal  to  two  right  angles.     §  18. 

(3)  If  from  a  point  any  number  of  straight  lines  are  drawn  in  a 

plane,  the  sum  of  all  the  angles  so  formed  is  equal  to  four 
right  angles.     §  18. 

(4)  All  straight  angles  are  equal.     §  21. 

(5)  All  right  angles  are  equal.     §  21. 

(6)  If  two  angles  are  equal,  their  supplements  are  equal,  and  two 

different  supplements  of  the  same  angle  are  equal.     §  21. 

(7)  If  two  adjacent  angles  are  supplementary,  their  outer  boun- 

daries lie  in  the  same  straight  line.     §  22. 

7.    On  the  Use  of  Instruments. 
See  §  26. 


CHAPTER   I 
TRIANGLES  AND  PARALLELOGRAMS 

Section  I 

GENERAL   PROPERTIES   OF   TRIANGLES 

32.  After  laying  so  much  of  a  foundation  we  are  ready  to 
proceed  with  the  task  of  building  up  the  main  body  of  geo- 
metrical knowledge,  and  it  must  be  constantly  borne  in  mind 
that  this  subject  is  a  continuous  development.  We  shall  meet 
new  truths  at  every  turn,  and  every  new  truth  will  aid  us  in 
the  establishment  of  still  other  new  truths. 

We  shall  first  present  a  few  easy  problems  connected  with 
the  construction  of  triangles. 

Definitions 

33.  A  triangle  whose  three  sides  are  unequal  in  length  is 
called  a  scalene  triangle. 

A  triangle  which  has  two  equal  sides  is  called  an  isosceles 
triangle. 

A  triangle  whose  three  sides  are  all  equal  is  called  an 
equilateral  triangle. 

Is  an  equilateral  triangle  also  isosceles  ? 
Is  an  isosceles  triangle  ever  equilateral  ? 
22 


32-34]  TRIANGLES  AND  PARALLELOGRAMS  23 


Proposition  I 

34.  It  is  required  to  construct  an  equilateral  triangle 
of  which  a  given  line-segment  is  one  side. 

First,  recall  what  an  equilateral  triangle  is.     If  one  side  is 

given  in  length,  how  long  are  the  other  two  sides  ? 
Now  state  the  proposition  in  specific  terms,  thus  : 


Let  AB  (see  the  diagram)  be  the  given  line-segment. 
It  is  required  to  construct  an  equilateral  triangle,  of  which 
AB  is  one  side. 

Next  proceed  with  the  construction. 

With  centre  A,  and  radius  AB,  describe  a  circle. 
With  centre  B,  and  radius  BA,  describe  a  circle. 
These    circles  will   intersect  in  two  points.     Why?     (See 
Art.  25.) 

Let  C  be  one  of  their  points  of  intersection. 
Join  AC  and  BC. 

Note.  When  we  say  'Join  ^C,'  we  mean  draw  a  straight  line  from 
the  point  A  to  the  point  C. 

The  triangle  ACB  so  constructed  has  the  line-segment  AB 
for  one  side ;  but  it  remains  to  show  that  this  triangle  is  equi- 
lateral. 

Proof.  Because  A  is  the  centre  of  the  first  circle,  and  the 
line-segments  AB  and  AC  are  radii,  therefore  AG  equals  AB. 


24  ELEMENTARY   GEOMETRY  [Chap.  I 

Because  B  is  the  centre  of  the  other  circle,  and  the  line- 
segments  BA  and  BG  are  radii,  therefore  BC  equals  BA. 

That  is,  AC  and  BC  are  each  equal  to  AB. 

Hence  AC  equals  BC.  (Axiom  1) 

Therefore  AB,  AC,  BC  are  all  equal,  and  the  triangle  ABC 
is  equilateral. 

EXERCISES 

1.  Is  this  proposition  a  problem  or  a  theorem  ? 

2.  If  F  is  the  other  point  of  intersection  of  the  two  circles,  and  we 
join  AF  and  BF,  prove  that  AFB  is  also  an  equilateral  triangle. 

3.  Join  CF.    Show  that  the  triangles  ACF  and  BCF  are  each  isosceles. 

35.  The  student  should  carefully  note  the  order  of  arrange- 
ment in  the  preceding  proposition.  The  same  order,  written 
out  in  less  detail,  perhaps,  will  be  followed  in  all  the  proposi- 
tions.    We  have 

1.  The  general  enunciation  of  the  proposition ;  that  is,  the 
statement  of  the  proposition  in  general  terms,  and  the  student, 
after  reading  the  enunciation,  should  proceed  no  further  until 
he  thoroughly  understands  just  w^hat  he  is  expected  to  do. 

2.  A  re-statement  of  the  proposition  applied  to  a  particular 
figure,  or  what  is  sometimes  called  the  particular  enunciation." 

3.  Making  the  necessary  construction. 

4.  Proof  that  the  thing  constructed  is  what  was  required. 
In  the  case  of  a  theorem  the  proof  goes  to  show  that  the  state- 
ment made  in  the  enunciation  is  true. 

36.  In  naming  a  circle  we  usually  mention  three  points  on 
it,  these  being  sufficient  to  distinguish  it  from  every  other  circle. 

Thus,  in  the  diagram  of  this  proposition,  the  circle  whose 
centre  is  A  would  be  called  the  circle  BCD,  and  the  circle 
whose  centre  is  B  would  be  called  the  circle  ACE. 

Exercise.     Would  naming  its  centre  and  one  point  on  it  be 
sufficient  to  distinguish  a  circle  from  all  other  circles  ? 


34-37]  TRIANGLES  AND   PARALLELOGRAMS  25 

Proposition  II 

[General  Enunciation] 

37.  To  construct  a  triangle  two  of  whose  sides  shall  he 
each  equal  to  one  given  line-segment,  and  the  third  side 
equal  to  another  given  line-segment. 

What  sort  of  a  triangle  will  this  be  ? 


[Particular  Enunciation] 

Let  m  and  n  be  the  given  line-segments. 

It  is  required  to  construct  a  triangle,  of  which  two  sides  are 
each  equal  to  m  and  the  third  side  equal  to  n. 

[Construction] 

Draw  anywhere  in  the  plane  a  line-segment  AB  equal  to  n. 
With  centre  A,  and  radius  equal  to  m,  describe  a  circle  CDF. 
With  centre  B,  and  radius  also  equal  to  m,  describe  a  circle 
CEF. 
Suppose  these  two  circles  intersect  at  C. 

Do  these  circles  necessarily  intersect  ?     If  so,  at  how  many 
points  ? 

Join  CA  and  CB. 

Then  CAB  is  the  triangle  required. 


26  ELEMENTARY  GEOMETRY  [Chap.  I 

[Proof] 

Let  the  pupil  make  a  diagram  for  himself  and  letter  it  as  in 
the  construction. 

In  the  triangle  CAB,  the  side  AC  is  a  radius  of  the  circle 
CDF,  hence  AC  equals  m. 

All  radii  of  a  circle  are  equal  by  definition.    (See  Art.  25.) 

The  side  BC  is  a  radius  of  the  circle  CEF,  hence  BC  equals 
m.     Explain  why. 

The  side  AB  was  constructed  equal  to  w. 

Hence  the  triangle  CAB  fulfils  the  given  conditions,  having 
two  sides  equal  to  m,  and  one  side  equal  to  n. 

EXERCISES 

1.  Do  the  two  circles  in  the  above  construction  necessarily  intersect  ? 
Suppose  the  line  n  were  more  than  twice  as  long  as  m,  what  then  ?  Can 
a  triangle  be  formed  with  sides  6  feet,  6  feet,  and  13  feet  in  length  ? 

2.  If  the  line-segments  m  and  n  are  equal,  what  sort  of  triangle  does 
CAB  become  ? 

3.  Construct  a  triangle  two  of  whose  sides  shall  be  each  equal  to  the 
longer  line-segment  n,  and  the  third  side  equal  to  the  shorter  line- 
segment  »n. 

4.  If  AF,  BF,  and  Ci^  are  joined,  show  that  ACF  and  BCF  are  each 
isosceles  triangles. 

5.  Show  how  to  find  a  point  equidistant  from  two  given  points, 
making  use  only  of  the  compasses. 

38.  For  convenience,  one  side  of  a  triangle  is  sometimes 
called  the  base  of  the  triangle,  and  the  other  two,  the  sides. 
The  base  may  be  any  side  whatever ;  the  opposite  angle  is  then 
called  the  vertical  angle,  and  the  vertex  of  that  angle,  the  vertex 
of  the  triangle.  The  two  angles  adjacent  to  the  base  are  called 
the  base  angles. 

In  an  isosceles  triangle  it  is  usually  the  unequal  side  which 
is  called  the  base. 


37-39]  TRIANGLES  AND   PARALLELOGRAMS  27 


Proposition  III 

39.  To  construct  a  triangle  having  its  sides  respectively 
equal  to  three  given  line-segments. 


Let  m,  n,  p,  be  the  three  given  line-segments. 

It  is  required  to  construct  a  triangle  whose  sides  are  respec- 
tively equal  to  m,  7i,  and  p. 

Construction.  Draw  a  line-segment  AB  anywhere  in  the 
plane  equal,  say,  to  m,  the  longest  of  the  three  given  line-seg- 
ments. 

With  centre  A,  and  radius  equal  to  n,  describe  a  circle 
CDE. 

With  centre  B,  and  radius  equal  to  p,  describe  a  circle 
CHF. 

Suppose  these  two  circles  intersect  at  C. 

Join  CA  and  CB. 

The  triangle  CAB  is  the  triangle  required. 

Proof.  In  the  triangle  CAB,  the  side  CA  equals  the  given 
line-segment  n  [why  ?] ;  the  side  CB  equals  the  given  line- 
segment  p  [why  ?]  ;  and  the  side  AB  was  chosen  equal  to  the 
given  line-segment  m. 

Hence  the  triangle  CAB  fulfils  the  given  conditions. 


28  ELEMEJNTARY   GEOMETRY  [Chap.  I 


EXERCISES 

1.  Under  what  conditions  would  the  circle  CDE  pass  through  the 
point  5,  or  the  circle  CHF  through  the  point  A  ? 

2.  Under  what  conditions  would  the  two  circles  CDE  and  CHF  not 
intersect  ?     Could  the  required  triangle  be  constructed  in  that  case  ? 

3.  If  the  line  m  is  greater  than  the  sum  of  n  and  p,  see  what  the  dia- 
gram would  look  like  if,  instead  of  choosing  the  first  line  equal  to  m,  you 
should  choose  it  equal  to  n  ov  p. 

4.  Can  you  infer  from  this  solution  that  one  side  of  a  triangle  cannot 
be  greater  than  the  sum  of  the  other  two  sides  ?  Keep  this  in  mind  for 
future  consideration. 


40.  If  two  triangles  are  identically  equal,  either  of  them  can 
be  made  to  occupy  the  position  of  the  other  (Art.  14) ;  or,  in 
other  words,  the  one  can  be  placed  on  the  other  so  that  their 
vertices  will  coincide  and  their  sides  coincide.  In  this  case  the 
three  sides  of  either  triangle  must  be  equal,  respectively,  to 
the  three  sides  of  the  other,  and  tlie  three  angles  of  either, 
equal  respectively  to  the  three  angles  of  the  other,  and  the 
two  triangles  are  said  to  be  equal  in  all  respects. 

But  the  question  arises  immediately,  If  two  triangles  are 
equal  in  some  respects,  is  not  this  sufficient  to  make  them 
equal  in  other  respects,  or  perhaps  in  all  respects  ? 

In  other  words,  if  two  triangles  can  be  made  to  coincide  in 
some  of  their  parts,  will  they  not  necessarily  coincide  in  all 
their  parts  ? 

We  speak  of  the  three  sides  and  the  three  angles  of  a  triangle 
as  the  parts  of  the  triangle. 

The  next  two  or  three  propositions  will  relate  to  questions 
of  this  kind. 

The  method  of  testing  the  equality  of  two  geometrical  figures 
by  placing  the  one  on  the  other  is  known  as  the  method  of 
superposition. 


39-41] 


TRIANGLES  AND  PARALLELOGRAMS 


29 


Proposition  IV 

41.  If  two  triangles  have  two  sides  and  the  included 
angle  of  the  one  equal,  respectively,  to  two  sides  and  the 
included  angle  of  the  other,  the  triangles  are  identi- 
cally equal. 


Let  ABC  and  DEF  be  the  two  given  triangles  in  which  the 
side  AB  equals  the  side  DE,  the  side  AC  equals  the  side  BF, 
and  the  angle  BAC  equals  the  angle  EDF. 

It  is  required  to  prove  that  the  third  side  BC  equals  the  third 
side  EF,  that  the  angle  ABC  equals  the  angle  DEF,  and  that 
the  angle  ACB  equals  the  angle  DFE. 

Proof.  Place  the  triangle  DEF  upon  the  triangle  ABC  so 
that  the  vertex  D  shall  coincide  with  the  vertex  A,  and  the 
side  DE  shall  fall  on  its  equal  side  AB. 

Then  because  DE  equals  AB,  E  must  coincide  with  B. 
Since  the  angle  EDF  equals  the  angle  BAC,  and  DE  coincides 
with  AB,  DF  will  fall  on  AC. 

It  may  be  necessary  to  turn  the  triangle  over  in  order  to  make 
BF  fall  on  AC,  as  would  be  the  case  with  the  triangle 
D'E'F',  but  this  is  allowable.     (See  Postulate  4.) 

And  because  DF  equals  AC,  F  must  coincide  with  C 

Call  to  mind  at  this  point  what  three  parts  of  one  triangle 
were  given  equal  to  three  parts  of  the  other. 

Now,  two  points  E  and  F  of  the  straight  line  EF  have  been 
made  to  coincide,  respectively,  with  two  points  B  and  C  of  the 
straight  line  BC. 


30  ELEMENTARY  GEOMETRY  [Chap.  I 

Hence  these  two  straight  lines  must  coincide  throughout. 

If  two  straight  lines  coincide  in  two  points,  ttiey  coincide  at 
every  point.     (Art.  4.) 

And  the  line-segment  EF  must  equal  the  line-segment  BC. 

Therefore,  since  the  two  triangles  ABC  and  DBF  have  been 
made  to  coincide  in  all  their  vertices  and  all  their  sides,  their 
angles  must  be  equal,  respectively,  and  the  two  triangles  must 
be  identically  equal. 

Notice  particularly  that  in  these  two  triangles  the  angles 
which  are  equal  are  opposite  sides  which  are  equal. 

Such  angles  or  sides  are  said  to  be  similarly  situated  in  the 
two  triangles.  They  are  sometimes  called  corresponding  angles 
or  sides ;  or  homologous  angles  or  sides. 

42.  In  a  theorem  there  are  always  two  distinct  parts.  It 
reads : 

If  certain  things  are  true,  the^i  certain  other  things  are  true. 
The  first  part,  the  z/part,  is  called  the  hypothesis.  The  second 
part,  the  then  part,  is  called  the  conclusion. 

The  theorem  of  Proposition  IV  may  be  divided  into  hypothe- 
sis and  conclusion,  thus : 

Hypothesis.  If  two  triangles  have  two  sides  and  the  in- 
cluded angle  of  the  one  equal,  respectively,  to  two  sides  and  the 
included  angle  of  the  other, 

Conclusion.     TJien  the  triangles  are  identically  equal. 

Separate  the  next  proposition  into  hypothesis  and  conclusion. 

EXERCISES 

In  the  two  triangles  ABC  and  DEF, 

1.  If  AB  equals  DE,  and  AC  equals  DF,  but  the  angle  BAC  is  greater 
than  the  angle  EDF,  where  would  DF  fall  when  the /triangle  DEF  is 
placed  upon  the  triangle  ABC  as  in  this  proposition  ? 

2.  If  the  angles  were  equal,  but  AC  greater  than  DF,  where  would 
F  fall  ? 

3.  Is  Proposition  IV  a  problem  or  a  theorem  ? 


41-43]  TRIANGLES  AND  PARALLELOGRAMS  31 


Proposition  V 

43.  If  two  triangles  have  a  side  and  the  two  adjacent 
angles  of  the  one  equal,  respectively,  to  a  side  and  the 
two  adjacent  angles  of  the  other,  the  triangles  are  iden- 
tically equal.* 


This  proposition,  like  the  last  one,  is  proved  by  superposition. 

Let  ABC  and  DEF  (or  D'E'F')  be  two  given  triangles  in 
which  the  side  EF  is  equal  to  the  side  BC,  and  the  angles  at 
E  and  F  are  respectively  equal  to  the  angles  at  B  and  C. 

It  is  required  to  prove  that  the  two  triangles  are  identically 
equal. 

Proof.  Place  the  triangle  DEF  upon  the  triangle  ABO  so 
that  the  side  EF  shall  coincide  with  its  equal  side  BC,  the 
point  E  coinciding  with  B,  and  F  with  C,  and  so  that  the  ver- 
tices D  and  A  lie  on  the  same  side  of  BC. 

Then,  since  the  angle  FED  equals  the  angle  CBA,  the  side 
ED  must  fall  on  BA,  and  the  point  D  will  coincide  with  some 
point  of  the  line  BA. 

Why  do  we  not  know  at  this  stage  that  D  coincides  with  A  ? 

Also,  since  the  angle  EFD  equals  the  angle  BCA,  the  side 
FD  must  fall  on  the  side  CA,  and  the  point  D  will  coincide 
with  some  point  of  the  line  CA. 

*  It  should  be  observed  that  by  '  the  two  adjacent  angles '  we  mean 
the  two  angles  adjacent  to  this  side,  or  of  which  this  side  is  a  common 
boundary.     Do  not  confuse  this  with  '  adjacent  angles '  defined  on  p.  8. 


32  ELEMENTARY  GEOMETRY  [Chap.  I 

Since  D  coincides  both  with  some  point  of  BA  and  with 
some  point  of  GA,  it  must  coincide  with  their  point  of  inter- 
section, namely,  with  A. 

Therefore,  since  the  two  triangles  have  been  made  to  coin- 
cide in  all  their  vertices  and  all  their  sides,  they  are  identically 
equal. 

44.  It  will  be  interesting  to  notice  the  relation  between  the 
last  two  propositions. 

Proposition  IV  may  be  stated  : 

Hypothesis.  If  two  triangles  have  two  sides  and  the  included  angle 
of  the  one  equal,  respectively,  to  two  sides  and  the  included  angle  of  the 
other, 

Conclusion.  Then  the  base  and  the  two  adjacent  angles  of  the  one  are 
equal,  respectively,  to  the  base  and  the  two  adjacent  angles  of  the  other. 

Proposition  V  may  be  stated: 

Hypothesis.  If  two  triangles  have  the  base  and  the  two  adjacent 
angles  of  the  one  equal,  respectively,  to  the  base  and  the  two  adjacent 
angles  of  the  other, 

Conclusion.  Then  the  two  sides  and  the  included  angle  of  the  one  are 
equal,  respectively,  to  the  two  sides  and  the  included  angle  of  the  other. 

Stated  in  this  way,  the  hypothesis  in  Proposition  IV  becomes 
the  conclusion  in  Proposition  V,  and  the  conclusion  in  Propo- 
sition IV,  the  hypothesis  in  Proposition  V. 

Two  theorems  so  related  are  said  to  be  converse. 

Definition.  Two  theorems  are  converse  when  the  hypothe- 
sis of  each  is  the  conclusion  of  the  other. 

Notation 

45.  Hereafter,  whenever  it  is  desirable,  we  shall  make  use 
of  certain  symbols  to  abbreviate  our  forms  of  statement.  For 
instance,  instead  of  writing  in  full  '  the  angle  ABC,'  we  shall 
write  Z  ABC,  instead  of  'the  angle  at  B,'  we  shall  write  Z5; 
instead  of  '  the  triangle  HKL,'  we  shall  write  A  HKL ;  and 
instead  of  '  is  equal  to,'  or  '  equals,'  we  shall  sometimes  use  the 


43-47]  TRIANGLES  AND  PABALLELOGRAMS  66 

algebraic  symbol  =.  But  care  should  be  taken  in  every  case 
to  read  the  symbol  as  though  the  statement  were  written  out 
in  full. 

On  Bisectors 

46.  If  through  the  vertex  of  an  angle  BAG,  we  draw  any 
straight  line  AD,  it  divides  the  angle  into  two  parts,  BAD 
and  DAC.  These  two  parts  may  be 
equal,  or  they  may  be  unequal.  If 
they  are  equal,  the  line  AD  is  said 
to  bisect  the  angle  BAC. 

If  the  angles  BAD  and  DAC  are 
unequal,  suppose  that  BAD  is  smaller 
than  DAC,  and  rotate  AD  in  the  way 
indicated   by   the   arrowhead.       This 

will  continuously  increase  the  one  angle  and  continuously  de- 
crease the  other,  while  it  leaves  their  sum  unaltered. 

It  is  clear  that  there  must  be  some  position  of  AD  for 
which  the  angles  BAD  and  DAC  are  equal ;  and  that  if  AD 
is  uioved  ever  so  little  from  that  position,  one  of  these  angles 
will  be  increased  while  the  other  is  diminished,  so  that  they 
will  become  unequal. 

Hence, 

Tlirough  the  vertex  of  any  angle  there  can  be  drawn  one,  and 
only  one,  straight  line  which  bisects  the  angle. 

If  the  angle  bisected  is  a  straight  angle  (see  Art.  20),  the 
bisector  is  perpendicular  to  the  straight  line  which  bounds  the 
angle.     From  this  it  follows  immediately  that 

At  any  point  in  a  straight  line  there  cari  be  drawn  one,  and 
only  one,  perpendicular  to  the  line. 

47.  Just  as  we  are  able  to  show  that  there  is  one,  and  only 
one,  'bisector'  of  an  angle,  so  we  might  show  that  there  is 
one,  and  only  one,  '  point  of  bisection '  or  '  mid-point '  of  a  line- 
segment. 


34  ELEMENTARY  GEOMETRY  [Chap.  I 

Proposition  VI 

48.   If  two  sides  of  a  triangle  are  equals  the  angles 
opposite  those  sides  are  also  equal. 

A 


Let  ABC  be  an  isosceles  triangle,  BA  and  CA  being  the 
equal  sides. 

It  is  required  to  prove  that  the  angle  at  (7,  opposite  the  side 
BA,  equals  the  angle  at  J5,  opposite  the  side  CA. 

Proof.  Suppose  that  the  line  AB  is  the  bisector  of  Z  BAC, 
and  that  it  meets  BC  at  Z).  It  thus  divides  A  BAC  into  two 
triangles  ;  namely,  A  BAT)  and  A  CAD. 

In  these  two  triangles,  the  sides  BA  and  AD  are  equal, 
respectively,  to  the  sides  CA  and  AD,  and  the  included  angle 
BAD  equals  the  included  angle  CAD. 

Hence  As  BAD  and  CAD  are  equal  in  all  respects. 

(Prop.  IV.) 

Therefore  Z  B  equals  A  C,  these  being  corresponding  angles, 
i.e.  opposite  equal  sides,  in  the  two  triangles. 

49.  The  proof  of  this  proposition  contains  the  proof  of  the 
following  theorem: 

Tlie  straight  line  which  bisects  the  vertical  angle  of  an  isosceles 
triangle  also  bisects  the  base,  and  is  perpendicular  to  the  base. 

For  since  As  BAD  and  CAD  are  identically  equal,  BD  =  CD 
and  Z  BDA  =  Z  CD  A.  Therefore  BC  is  bisected  at  the 
point  D,  and  AD  is  perpendicular  to  BC. 


48-50]  TRIANGLES  AND  PARALLELOGRAMS  35 

50.  Definition.  A  theorem,  the  truth  of  which  is  easily 
deduced  from  another,  or  the  proof  of  which  is  easily  obtained 
from  the  proof  of  another,  is  said  to  be  a  corollary  of  that 
other. 

The  following  is  a  second  corollary  to  Proposition  VI : 

Corollary.     An  equilateral  triangle  is  also  equiangular. 

The  pupil  should  show  that  if  the  three  sides  of  a  triangle  are 
equal,  the  three  angles  are  equal. 

While  the  proof  of  a  corollary  is  easily  deduced  from  that 
of  the  main  proposition,  it  should  never  be  neglected,  but 
should  in  most  cases  be  carefully  written  out  by  the  pupil. 


EXERCISES 

1.  If  two  isosceles  triangles,  ABC  and  DBO,  are  on  the  same  base, 
BC,  but  on  opposite  sides  of  it,  prove  (1)  that  the  angle  ABD  is  equal 
to  the  angle  ACD,  (2)  that  if  AD  be  drawn  A  ABD  is  identically- 
equal  to  AACD,  (3)  that  AD  bisects  the  vertical  angle  of  each  triangle, 
(4)  that  AD  bisects  the  common  base  of  the  triangles,  (5)  that  AD 
makes  right  angles  with  the  base. 

2.  Prove  the  same  things  when  the  two  triangles  lie  on  the  same  side 
of  the  base. 

3.  If  the  vertex  of  an  isosceles  triangle  is  joined  to  the  mid-point  of 
the  base,  prove  that  the  line  so  drawn  bisects  the  vertical  angle  of  the 
triangle,  and  also  that  it  is  perpendicular  to  the  base. 

4.  On  a  given  line-segment  construct  two  different  isosceles  triangles, 
and  make  use  of  these  to  find  the  mid-point  of  the  given  line-segment. 

5.  Prove  that  the  triangle  whose  vertices  are  the  mid-points  of  the 
sides  of  an  equilateral  triangle  is  equilateral.     (Apply  Prop.  IV.) 

6.  Prove  that  the  triangle  whose  vertices  are  the  raid-points  of  the 
sides  of  an  isosceles  triangle  is  isosceles.     (Apply  Prop.  IV.) 

7.  If  on  the  sides  AB,  BC,  CA,  of  an  equilateral  triangle  ABC,  equal 
lengths  AP^  BQ^  CB,  are  taken,  prove  that  the  triangle  PQB  is  also 
equilateral. 


36  ELEMENTARY  GEOMETRY  [Chap.  I 


Proposition  YII 

51.   If  two  angles  of  a  triangle  are  equal,  the  sides 
opposite  those  angles  are  also  equal. 


Let  ABC  be  a  triangle  in  which  the  angle  ABC  equals  the 
angle  ACB. 

It  is  required  to  prove  that  the  side  CA  is  equal  to  the 
side  BA. 

Proof.  Mark  the  positions  of  the  vertices  and  sides  of  the 
triangle,  and  then  suppose  the  triangle  to  be  turned  over  and 
put  down  so  that  the  vertex  B  takes  the  former  position  of  the 
vertex  C,  and  the  vertex  C  the  former  position  of  the  vertex  B. 

The  side  BC  will  thus  occupy  its  former  position,  but  will 
be  turned  end  for  end. 

Since  Z  B  equals  Z  C,  the  side  BA,  when  the  triangle  is 
turned  over,  will  fall  upon  the  former  position  of  the  side 
CA,  and  the  side  CA  will  fall  upon  the  former  position  of 
the  side  BA. 

Therefore  the  point  A  will  occupy  its  former  position,  and 
the  side  BA  will  coincide  with  the  former  position  of  the 
side  CA. 

Therefore  the  side  BA  equals  the  side  CA. 

52.   Corollary.     An  equiangular  triangle  is  also  equilateral. 

The  pupil  should  prove  that  if  the  three  angles  of  a  triangle 
are  equal,  the  three  sides  must  be  equal. 


51-53] 


TRIANGLES  AND  PARALLELOGRAMS 


37 


Proposition  VIII 

53.  //  two  triangles  have  the  three  sides  of  the  one 
equal,  respectively,  to  the  three  sides  of  the  other,  the  two 
triangles  are  identically  equal. 


Let  ABC  and  DEF  be  two  triangles  having  the  side  AB 
equal  to  the  side  DE,  BC  equal  to  EF,  and  CA  equal  to  FD. 

It  is  required  to  prove  that  the  two  triangles  are  identically 
equal. 

Proof.  Place  the  triangle  DEF  so  that  the  side  EF  will 
coincide  with  its  equal  side  BC,  the  point  E  with  B,  and  the 
point  F  with  C,  and  so  that  the  vertex  D  will  fall  on  the  oppo- 
site side  of  BC  from  the  vertex  A. 

Join  AD. 


Then  in  A  ABD,  Z  BAD  =  Z  BDA, 
and  in         A  ACD,  Z  CAD  =  Z  CDA. 


Therefore 


ZBAC=ZBDa 


(Prop.  VI.) 

Why? 

(Axiom  2.) 


But  Z  BDC  is  Z  EDF,  which  is  therefore  equal  to  Z  BAC. 

Hence  in  As  BAC  and  EDF,  two  sides  and  the  included 
angle  of  one  are  equal,  respectively,  to  two  sides  and  the  in- 
cluded angle  of  the  other.     Name  them. 

Therefore  the  two  triangles  are  identically  equal.    (Prop.  IV.) 


38 


ELEMENTARY  GEOMETRY 


[Chap.  I 


In  Article  46  we  showed  that  for  every  angle  there  is  one 
and  only  one  bisector,  and  also  stated  in  Art.  47  that  for  every 
line-segment  there  is  one  and  only  one  mid-point,  or  bisecting 
point.  In  the  next  two  propositions,  we  shall  give  simple 
methods  for  finding  these  bisectors  with  the  ruler  and 
compasses. 

Proposition  IX 

54.   To  bisect  a  given  angle. 


Let  BAC  be  a  given  angle. 

It  is  required  to  draw  the  straight  line  which  will  bisect  it. 


From  AB  and  AC  cut  off  any  equal  segments 
[Use  compasses.] 


Construction. 
AD  and  AE. 

Join  DE. 

Upon  DE  as  base  construct  any  isosceles  triangle  having  its 
vertex  F  on  the  opposite  side  of  DE  from  A.  (Prop.  II.) 

Join  AF. 

Then  AF  is  the  required  line. 

Proof.     Compare  As  ADF  and  AEF  side  for  side. 
These  triangles  are  equal  in  all  respects.     Where  proved  ? 
Therefore  Z  DAF  equals  Z  EAF.    That  is  Z  BAC  is  bisected 
by  .4i^. 


53-55]  TRIANGLES  AND  PARALLELOGRAMS  39 

55.  It  should  be  noticed  that  in  order  to  draw  the  bisector 
AF,  it  is  not  necessary  to  draw  the  straight  lines  DE,  DF,  and 
EF.  The  points  D,  E,  and  F  can  be  located  with  the  com- 
passes, and  then  the  line  AF  drawn,  and  in  actual  practice 
this  is  all  that  should  be  done.  The  other  lines,  however,  are 
necessary  for  the  proof. 

EXERCISES 

1.  If  ABC  is  an  isosceles  triangle,  and  the  equal  angles  ABC  and 
ACB  are  bisected  by  the  lines  BD  and  CD  which  meet  at  i>,  prove  that 
DBC  is  also  an  isosceles  triangle. 

2.  ABC  is  an  isosceles  triangle  having  AB  equal  to  AC.  In  AB  and 
AC  tvs^o  points  D  and  E  are  taken  equally  distant  from  A  ;  prove  that 
the  line-segments  BE  and  CD  are  equal.  (Prop.  IV.) 

3.  If  two  line-segments  bisect  each  other  at  right  angles,  prove  that 
any  point  in  one  of  them  is  equidistant  from  the  extremities  of  the  other. 

4.  BAC  is  a  triangle  having  the  angle  B  double  of  the  angle  A  ;  if  BD 
bisects  the  angle  B  and  meets  J  C  at  Z>,  prove  that  BD  equals  AD. 

5.  ACB  and  ADB  are  two  triangles  on  the  same  side  of  AB.,  such  that 
AC  IS  equal  to  BD.,  and  AD  is  equal  to  BC,  AC  and  BD  intersecting 
at  0.     Show  that  the  triangle  AOB  is  isosceles. 

6.  Divide  a  given  angle  into  four  equal  parts. 

7.  If  in  a  triangle  DEF  the  angles  D  and  E  are  bisected  by  straight 
lines  meeting  at  H,  and  Di/ equals  EH.,  prove  that  Z)F  equals  EF. 

8.  Prove  that  the  bisectors  of  the  angles  of  an  equilateral  triangle 
meet  in  one  point.     Is  the  same  thing  true  of  an  isosceles  triangle? 

Suggestion.  Draw  the  bisectors  of  two  equal  angles  and  let  them 
meet  at  0.  Join  0  to  the  third  vertex,  and  show  that  the  line  so  drawn 
bisects  the  third  angle. 

9.  Apply  the  construction  of  Proposition  IX  to  bisect  a  straight  angle, 
i.e.  to  draw  a  perpendicular  to  a  given  straight  line  at  a  given  point. 

10.  If  D  and  E  are  the  mid-points  of  the  equal  sides  AC  and  AB  of  an 
isosceles  triangle  ABC^  prove  that  BD  equals  CE. 

11.  Let  D  be  any  point  on  the  bisector  of  an  angle  BA  C ;  prove  that  if 
AB  equals  AC,  the  angle  ADB  equals  the  angle  ADC. 


40 


EL EMEN  TA  R  Y  GEOMETR  Y 


[Chap.  1 


Proposition  X 
56.   To  bisect  a  given  line-segment. 


)^ 


y 


^< 


\. 


N. 


E 


y:D 


V 


/ 


T^ 


Let  AB  be  a  given  line-segment,  which  it  is  required  to 
bisect. 

Construction.  With  centre  A  and  any  convenient  radius 
describe  a  circle. 

With  centre  B,  and  an  equal  radius,  describe  another  circle. 
Let  these  circles  intersect  at  points  C  and  D. 

Note.  The  radius  must  be  chosen  long  enough  to  make  the  two 
circles  intersect. 

Draw  the  straight  line  CD,  meeting  AB  at  E. 
Then  E  is  the  mid-point  of  the  line-segment  AB. 

Proof.     Join  CA  and  CB,  also  DA  and  DB. 
Since  these  line-segments  are  all  equal.  As  ACD  and  BCD 
are  identically  equal.  (Prop.  VIII.) 

Therefore  Z  ACD  =  Z  BCD. 

In  the  triangles  ACE  and  BCE,  two  sides  and  the  included 
angle  in  one  are  equal  to  two  sides  and  the  included  angle  in 
the  other. 

Therefore  these  triangles  are  identically  equal.     (Prop.  IV.) 
And  the  line-segment  AB  is  bisected  at  the  point  E. 


56-67] 


TRIANGLES  AND  PARALLELOGRAMS 


41 


Since  CD  not  only  bisects  AB  at  E,  but  is  also  perpendicular 
to  AB,  it  may  be  called  the  perpendicular  bisector  of  AB. 


57.  Corollary.  Two  points  each  equidistant  from  the  ex- 
tremities of  a  given  line-segment  determine  the  perpendicular 
bisector  of  the  segment. 


/ 


/ 


\ 


\ 


E 
Fig.  1. 


The  figure  of  the  proposition  illustrates  one  case  of  the  theorem  stated 
in  the  corollary.  Another  case  arises  when  C  is  equidistant  from  A  and 
B^  and  also  D  equidistant  from  A  and  B^  as  in  Fig.  1. 

Suggestion.  It  is  here  required  to  prove  that  DC  bisects  AB  at 
right  angles.     From  As  ^Z>C  and  BDC  prove  ZADC=  /.BDC. 

(Prop.  VIII.) 
From  As  ADE  and  BDE  prove  AE  =  BE  and  Z  AED  =  Z  BED. 

(Prop.  IV.) 

EXERCISES 

1.  What  lines  of  the  figure  of  Proposition  X  were  necessary  to  find 
the  point  E,  and  what  lines  had  to  be  added  for  the  proof  ? 

2.  In  the  figure  of  Proposition  X  show  that  CD  is  also  bisected  at 
right  angles  by  AB. 

3.  If  two  circles  intersect,  the  straight  line  joining  their  centres  bisects 
at  right  angles  the  line-segment  joining  their  points  of  intersection. 

4.  Find  a  line-segment  equal  to  (1)  half  the  sum  of  two  given  line- 
segments,  (2)  half  the  difference  of  two  given  line-segments. 

6.   Find  a  line-segment  half  as  long  again  as  a  given  line-segment. 
6.    Prove  that  every  point  on  the  perpendicular  bisector  of  a  line- 
segment  is  equidistant  from  the  extremities  of  the  line-segment. 


42  ELEMENTARY  GEOMETRY  [Chap.  I 


Proposition  XI 

58.  If  two  straight  lines  intersect,  the  vertical  angles 
are  equal. 


Let  the  straight  lines  AB  and  CD  intersect  at  0. 
It  is  required  to  prove  that  the  vertical  angles  AOC  and  BOD 
are  equal. 

Proof.     Zs^OC  and  AOD  are  together  equal  to  two  right 
angles.  (Art.  16.) 

Hence  Z  AOC  is  the  supplement  of  Z  AOD. 

The  Zs  BOD  and  AOD  are  together  equal  to  two  right  angles. 
Why? 

Hence  Z  BOD  is  the  supplement  of  Z  AOD. 

Therefore  Z  AOC  and  Z  jBOZ>,  being  supplements  of  the 
same  angle,  are  equal.  (Art.  21.) 

Similarly  it  may  be  shown  that  the  vertical  angles  AOD  and 
BOC  are  equal. 

EXERCISES 

1.  The  straight  line  which  bisects  one  of  two  vertical  angles  bisects 
also  the  other. 

2.  The  straight  lines  which  bisect  two  adjacent  supplementary  angles 
are  at  right  angles  to  each  other. 

3.  If  one  straight  line  is  perpendicular  to  a  second,  the  second  is  per- 
pendicular to  the  first. 

In  connection  with  this  example  you  will  need  carefully  to  recall  the 
definition  of  a  perpendicular  given  in  Art.  15. 

4.  If  one  of  the  four  angles  made  by  two  intersecting  straight  lines  is  a 
right  angle,  the  other  three  are  also  right  angles. 


68-59]  TRIANGLES  AND   PARALLELOGRAMS  43 


Proposition  XII 

59.  If  any  side  of  a  triangle  is  produced,  the  exterior 
angle  so  formed  is  greater  than  either  of  the  two  interior 
non-adjacent  angles. 


Let  the  triangle  ABC  have  the  side  BC  produced  to  D. 

It  is  required  to  prove  that  the  angle  ACD  is  greater  than 
either  the  angle  at  A  or  the  angle  at  B. 

Proof.  Bisect  the  side  AC  at  E,  join  BE,  and  produce  to 
F,  making  EF  equal  to  BE.     Join  FC 

A  EEC  is  identically  equal  to  A  EBA.     Why  ? 

Apply  Propositions  XI  and  IV. 

Therefore  Z  ECF  equals  its  corresponding  Z  EAB. 
But  Z  ACD  is  greater  than  Z  ECF.      Therefore  Z.ACD  is 
greater  than  Z  CAB,  or  Z.A. 

To  prove  Z  ACD  greater  than  Z  B,  it  is  only  necessary  to 
produce  the  side  AC,  thus  forming  a  vertical  angle  equal  to 
Z  ACD,  then  proceed  just  as  before,  bisecting  the  side  BC 
instead  of  the  side  AC. 

EXERCISES 

1.  Show  that  ZA  is  less  than  ZAEF,  and  that  Zi^  is  less  than 
ZBEC. 

2.  Draw  three  figures  to  show  that  an  exterior  angle  of  a  triangle  may 
be  greater  than,  equal  to,  or  less  than  the  interior  adjacent  angle. 


44  ELEMENT ABY  GEOMETRY  [Chap.  I 

Proposition  XIII 

60.  Any  two  angles  of  a  triangle  are  together  less 
than  two  right  angles. 

Suggestions  for  Proof.  Produce  one  side  of  the  triangle 
and  so  form  an  exterior  angle. 

This  exterior  angle  is  greater  than  either  of  the  interior  non- 
adjacent  angles.  (Prop.  XII.) 

The  sum  of  the  interior  adjacent  angle  and  this  exterior 
angle  is  equal  to  two  right  angles.  (Art.  16.) 

Therefore  the  sum  of  the  interior  adjacent  angle  and  either 
of  the  other  interior  angles  is  less  than  two  right  angles. 

By  producing  a  different  side  of  the  triangle  the  angles  can 
be  taken  in  pairs  in  other  ways,  and  the  theorem  can  thus  be 
proved  for  all  possible  pairs. 

The  pupil  should  make  his  own  diagram  and  write  out  the 
proof  in  full,  naming  the  angles  used. 

61.  Corollary  I.  If  a  triangle  has  one  light  angle  or  one 
obtuse  angle,  the  other  two  angles  must  he  acute. 

Definition.  A  triangle  which  has  a  right  angle  is  called 
a  right  triangle. 

The  side  opposite  the  right  angle  is  called  the  hypotenuse. 

62.  Corollary  II.  From  a  point  outside  a  given  straight 
line,  not  more  than  one  perpendicular  to  the  line  can  he  drawn. 

EXERCISES 

1.  Show  by  joining  the  vertex  yl  of  a  triangle  ABC  with  any  point  of 
the  opposite  side  between  B  and  C  that  the  angles  at  B  and  C  are  together 
less  than  two  right  angles. 

2.  If  any  side  of  a  triangle  is  produced  both  ways,  the  two  exterior 
angles  so  formed  are  together  greater  than  two  right  angles. 

3.  The  angles  at  the  base  of  an  isosceles  triangle  are  both  acute. 


60-63]  TRIANGLES  AND  PARALLELOGRAMS  45 

Proposition  XIV 

63.  If  two  sides  of  a  triangle  are  unequal,  the  angle 
opposite  the  greater  side  is  greater  than  the  angle  oppo- 
site the  less. 


Let  ABC  be  a  triangle  in  which  the  side  AC  is  greater  than 
the  side  AB. 

It  is  required  to  prove  that  the  angle  B  is  greater  than  the 
angle  C. 

Proof.     From  AC  cut  off  a  part  AD  equal  to  AB.     Join  BD. 
The  line  BD  necessarily  falls  within  Z  ABC. 
In  A  ABD,  Z  ABD  equals  Z  ADB.     Why  ? 
But  Z  ^Z)J3  is  greater  than  Z  ACB.     Why  ? 
Therefore  Z  ^J5i)  is  greater  than  Z  ^C^. 
But  Z  ABC  is  greater  than  Z  ABD.  (Axiom  8.) 

Therefore  Z  ABC  is  greater  than  Z  ACB.  (Axiom  9.) 

EXERCISES 

1.  The  angles  in  a  scalene  triangle  are  all  unequal. 

2.  If  one  side  of  a  triangle  is  less  than  another,  the  angle  opposite  it 
must  be  acute. 

3.  Prove  the  proposition  by  producing  AB  to  D,  making  AD  equal  to 
AC  and  joining  DC. 

4.  If  straight  lines  are  drawn  from  any  point  within  a  triangle  to  two 
of  the  vertices  of  the  triangle,  the  angle  contained  by  these  lines  is  greater 
than  the  angle  at  the  third  vertex. 

Suggestion.  Draw  a  line  from  the  third  vertex  through  the  chosen 
point.     Then  compare  angles.     Find  also  a  second  proof. 


46  ELEMENT AEY  GEOMETBY  [CBur.  I 

64.   Axiom  10.     Two  magjiitudes  A  and  B  of  the  same  kind 
are  either  equal,  or  A  is  greater  than  B,  or  A  is  less  than  B, 


Proposition  XV 

65.  If  two  angles  of  a  triangle  are  unequal,  the  side 
opposite  the  greater  angle  is  greater  than  the  side  oppo- 
site the  less. 


Let  ABC  be  a  triangle  in  which  the  angle  B  is  greater  than 
the  angle  C. 

It  is  required  to  prove  that  the  side  AC  is  greater  than  the 
side  AB. 

Proof.  If  the  side  AG  is  not  greater  than  AB,  it  must  be 
equal  to  AB,  or  less  than  AB.  (Axiom  10. 

If  AC  were  equal  to  AB,  how  would  Zs  J5  and  C  compare  ? 
Give  reference. 

If  AC  were  less  than  AB,  how  would  Zs  B  and  C  compare  ? 
Give  reference. 

Conclusion. 

66.  Corollary  I.  In  a  right  triangle  the  hypotenuse  is  the 
greatest  side. 

67.  Corollary  II.  Not  more  than  two  equal  straight  lines 
can  be  drawn  from  a  giveri  point  to  a  given  straight  line. 

Suppose  a  third  line  AD,  equal  to  the  first  two,  AB  and  AC,  can  be 
drawn.    Then  ZABC=ZACB,  ZABD  =  ZADB,  and  ZACD=ZADC. 

(Prop.  VI.)     But  this  is  impossible  by  Proposition  XII. 


64-69]  TRIANGLES  AND  PARALLELOGRAMS  47 

68.  The  theorems  of  Propositions  XIV  and  XV  are  converse, 
as  are  also  those  of  Propositions  VI  and  VII. 

Proposition  VIII  may  be  stated : 

If  two  triangles  have  the  three  sides  of  the  one  equal,  respectively,  to 
the  three  sides  of  the  other,  then  the  three  angles  of  the  one  are  equal, 
respectively,  to  the  three  angles  of  the  other. 

S^e  the  converse.  Is  the  converse  true  ?  Can  a  theorem 
be  true  and  its  converse  be  false  ? 

69.  Special  attention  should  be  paid  to  the  method  of  proof 
in  Proposition  XV.  It  is  a  method  very  frequently  used 
when  we  wish  to  prove  the  converse  of  a  theorem  whose  truth 
has  already  been  established. 

In  the  theorem  of  this  proposition  the  statement  to  be  proved 
is  that  one  side  p  of  a  triangle  is  greater  than  another  side  q, 
upon  the  hypothesis  that  the  opposite  angle  P  is  greater  than 
the  opposite  angle  Q,  and  we  proceed  as  follows : 

1.  Suppose  that  the  statement  is  not  true.  The  only  other  possibili- 
ties are :  First,  p  is  equal  to  g,  or  second,  p  is  less  than  q. 

2.  Show  by  reference  to  a  previous  theorem  (Prop.  VI.)  that  if  p  is 
equal  to  q,  then  the  hypothesis  cannot  be  true. 

3.  Show  also  by  reference  to  a  previous  theorem  (Prop.  XIV.)  that  if  j? 
is  less  than  g,  then  again  the  hypothesis  cannot  be  true. 

4.  Conclude  that  since  with  the  given  hypothesis  neither  of  the  other 
possibilities  can  be  true,  the  statement  of  the  theorem  must  be  true. 

Care  must  be  taken  in  such  a  proof  that  all  possible  cases 
have  been  examined  before  drawing  a  conclusion. 
Such  a  proof  is  called  an  indirect  proof. 

EXERCISES 

1.  Prove  that  if  D  be  any  point  in  the  base  BC  of  an  isosceles  triangle 
ABC,  between  5  and  C,  AD  is  less  than  AB ;  but  if  D  lie  on  the  base 
produced  either  way,  then  AD  is  greater  than  AB. 

2.  ABC  is  a  triangle  in  which  OB  and  OC  bisect  the  angles  B  and  C, 
respectively  ;  show  that  if  AB  is  greater  than  AC,  then  OB  is  greater 
than  OC. 


48  ELEMENTARY  GEOMETRY  [Chap.  I 

Propositiox  XVI 

70.  tdny  two  sides  of  a  triangle  are  together  greater 
than  the  third  side. 


Let  ABC  be  any  triangle. 

It  is  required  to  prove  that  any  two  sides  AB  and  BC  are 
together  greater  than  the  third  side  AC. 

Proof.     Produce  the  side  AB  to  D,  making  BD  equal  to  BC. 

Join  DC. 

Then  AD  is  the  sum  of  AB  and  BC,  and  it  is  required  to 
prove  that  AD  is  greater  than  AC. 

Z  BCD  equals  Z  BDC     Why  ? 

Therefore  Z  ACD  is  greater  than  Z  BDC,  or  Z  ^Z)a 

Therefore  in  A  ADC,  the  side  ^Z>  is  greater  than  the  side 
AC.     Give  reference. 

Hence  in  A  ABC,  the  sides  AB  and  BC  are  together  greater 
than  the  side  AC. 

EXERCISES 

1.  Prove  this  theorem  by  bisecting  one  angle  and  producing  the  bisector 
to  meet  the  opposite  side. 

2.  Could  a  triangle  be  formed  whose  sides  are  7,  11,  and  19  feet, 
respectively  ?  What  is  the  limitation  on  the  construction  in  Proposition 
III. 

3.  Prove  that  the  difference  between  any  two  sides  of  a  triangle  is  less 
than  the  third  side. 

4.  A  side  of  an  isosceles  triangle  is  always  greater  than  half  the  base. 


70-71]  TRIANGLES  AND   PARALLELOGRAMS  49 

Proposition  XVII 

71.   Fi^om  a  given  point  in  a  straight  line  to^  draw  a 
perpendicular  to  the  line. 


/ 


/ 


A  E  C  D  B 

Let  AB  be  the  given  straight  line  and  C  the  given  point  in  it. 

It  is  required  to  draw  from  C  a  straight  line  which  is  per- 
pendicular to  AB. 

Construction.  From  C,  on  opposite  sides,  mark  off  any  equal 
segments  CD  and  CE. 

Upon  the  line-segment  ED  as  base,  construct  any  isosceles 
triangle  whose  vertex  is  F.     Draw  the  straight  line  CF. 

Then  CF  is  perpendicular  to  AB. 

Proof.  Compare  As  FCE  and  FCD,  making  use  of  Propo- 
sition VIII.     The  triangles  are  identically  equal. 

Hence  Z  FCE  equals  Z  FCD  and  each  of  them  is  a  right 
angle. 

Therefore  CF  is  the  required  line. 

Compare  the  proof  of  this  proposition  with  Ex.  9,  p.  39. 
For  a  proof  that  not  more  than  one  perpendicular  can  be  drawn 
from  any  point,  see  Article  6 

EXERCISES 

1.  Repeat  the  above  construction,  using  the  compasses  and  ruler, 
drawing  only  such  arcs  and  lines  as  are  necessary. 

2.  In  the  figure  above,  show  that  any  point  in  the  straight  line  CF, 
produced  ever  so  far  either  way,  is  equidistant  from  D  and  E^  while  a 
point  of  the  plane,  not  on  CF^  is  not  equidistant  from  D  and  E. 


50  ELEMENTARY  GEOMETRY  [Chap.  I 


On  Loci 

72.  If  you  should  be  given  the  problem  to  find  a  point  in 
a  plane  equidistant  from  two  given  points  in  that  plane,  you 
would  at  once  recognize  that  there  are  many  points  which  will 
satisfy  this  condition.  In  fact,  every  point  of  a  particular 
straight  line  fulfils  the.  condition,  while  no  point  off  that  line 
does  so.     (Ex.  2;  p.  49.) 

This  particular  straight  line  is  called  the  locus  of  the  points 
fulfilling  the  given  conditions. 

The  locLis  of  points  in  a  plane  which  are  at  a  given  distance 
from  a  given  point  is  a  circle  of  which  the  given  point  is  the 
centre,  and  the  given  distance  equal  to  a  radius,  since  all  points 
of  the  circle  are  at  the  given  distance  from  the  given  point, 
and  there  is  no  other  point  of  the  plane  which  is  at  that  dis- 
tance from  the  given  point. 

73.  Definition.  If  every  point  of  a  line  or  group  of  lines 
satisfies  given  conditions,  while  no  other  point  satisfies  those 
conditions,  that  line  or  group  of  lines  is  called  the  locus  of 
the  points  satisfying  the  given  conditions. 

If  any  moving  point  continues  to  satisfy  the  given  conditions, 
it  is  evident  that  it  must  follow  the  locus. 

74.  The  properties  of  loci  can  be  utilized  for  the  solution  of 
man}^  problems  in  geometry,  for  example  the  problem,  —  to 
find  a  point  which  is  equidistant  from  three  given  points  A,  B, 
and  a 

The  locus  of  points  equidistant  from  A  and  B  is  the  perpen- 
dicular bisector  of  the  line-segment  AB. 

The  locus  of  points  equidistant  from  B  and  C  is  the  per- 
pendicular bisector  of  the  line-segment  BC. 

If  these  two  perpendicular  bisectors  m  and  n  intersect  at  O, 
then  0  is  equidistant  from  Aj  B,  and  C,  and  is  the  only  such 
point. 


72-74]  TRIANGLES  AND   PARALLELOGRAMS  51 

For, 

1.  Every  point  of  m  is  equidistant  from  A  and  B ; 
Every  point  of  n  is  equidistant  from  B  and  C. 

0  is  a  point  of  both  m  and  7i,  and  is  therefore  equidistant 
from  A,  B,  and  C. 

2.  No  point  outside  of  m  is  equidistant  from  A  and  B ; 
No  point  outside  of  n  is  equidistant  from  B  and  C. 
Therefore,  no  point  other  than  their  intersection,  can  be 

equidistant  from  A,  B,  and  C. 

if  m  and  n  do  not  intersect,  there  is  no  point  equidistant 
from  A,  B,  and  (7. 

EXERCISES 

1.  If  Unes  are  drawn  to  the  extremities  of  a  given  line-segment  from 
any  point  of  its  perpendicular  bisector,  these  lines  are  not  only  equal  but 
make  equal  angles,  both  with  the  given  line-segment  and  with  its  perpen- 
dicular bisector. 

2.  Find  a  point  in  a  given  straight  line  such  that  its  distance  from  two 
given  points  may  be  equal. 

3.  If  from  any  point  within  a  triangle  line-segments  are  drawn  to  the 
extremities  of  one  side,  these  are  together  less  than  the  other  two  sides  of 
the  triangle. 

Suggestion.  Produce  one  of  these  line-segments  backward  to  meet 
a  side  of  the  triangle,  then  apply  Proposition  XVI  twice. 

Notice  what  was  proved  concerning  lines  so  drawn  in  Ex.  4,  p.  45. 

4.  If  0  be  any  point  within  a  triangle  ABC,  show  that  the  sum  of  OA, 
OB,  and  OC  is  less  than  the  sum  of  the  sides  of  the  triangle,  but  greater 
than  half  the  sum  of  the  sides, 

5.  The  segment  of  a  straight  line  between  any  two  points  is  shorter 
than  any  broken  line  connecting  those  points. 

6.  Find  a  point  which  is  equidistant  from  two  given  points  and  at  a 
given  distance  from  a  fixed  point. 

How  many  such  points  can  there  be  ?  Is  it  always  possible  to  find  a 
point  which  will  satisfy  the  given  conditions  ? 


52  ELEMENTABY  GEOMETRY  [Chap.  I 


Proposition  XVIII 

75.  From  a  given  point  without  a  given  straight  line 
to  draw  a  perpendicular  to  the  line. 


m 

« 

A 

o'\           E          yh 

B- -'' 

Let  m  be  the  given  straight  line  and  A  the  given  point. 

It  is  required  to  draw  from  A  a  straight  line  perpendicular 
to  m. 

Construction.  Choose  a  point  B  on  the  opposite  side  of  m 
from  A.  With  radius  AB  describe  a  circle  which  must  inter- 
sect m  in  two  points,  say  C  and  D.     (See  Art.  25.) 

Bisect  the  line-segment  CD  at  E.     Join  AE. 

Then  AE  is  the  required  perpendicular. 

Proof.     Compare  As  AEC  and  AED,  using  Proposition  VIII. 

The  triangles  are  identically  equal,  hence  Z  AEO  equals 
Z  AED,  and  therefore  AE  is  perpendicular  to  m. 

The  pupil  should  repeat  this  construction,  drawing  only  such 
lines  as  are  necessary  to  find  the  point  E. 

76.  It  was  shown  on  p.  44  that  from  a  point  A  there  could 
not  be  drawn  two  perpendiculars  to  the  same  straight  line; 
that  there  can  be  one  perpendicular,  we  have  here  shown  by 
actually  constructing  it.     Hence  the  theorem: 

From  a  point  outside  a  straight  line,  there  can  he  draivn  one 
perpendicular  to  that  line,  and  only  one. 

The  point  E  in  the  diagram  is  called  the  foot  of  the  perpen- 
dicular drawn  from  the  point  A  to  the  line  m. 


75-77]  TRIANGLES  AND  PARALLELOGRAMS  53 

77.  By  the  distance  from  a  point  to  a  straight  line,  we  mean 
the  length  of  the  line-segment  between  the  given  point  and  the 
foot  of  the  perpendicular  drawn  from  the  point  to  the  line.  In 
other  words,  the  distance  from  a  point  to  a  straight  line  is 
measured  along  the  perpendicular  from  the  point  to  the  line. 


EXERCISES 

1.  Of  all  the  line-segments  which  can  be  drawn  from  a  given  point 
to  a  given  straight  hne  not  passing  through  it,  the  perpendicular  is  the 
shortest. 

2.  From  a  given  point  line-segments  are  drawn  to  a  given  straight 
line,  making  equal  angles  with  the  perpendicular,  on  opposite  sides  ;  prove 
that  these  line-segments  are  equal  and  intercept  equal  segments  of  the 
given  line  on  opposite  sides  of  the  perpendicular  ;  and,  conversely,  if  they 
intercept  equal  segments  of  the  line,  then  they  are  equal  and  make  eqife-l 
angles  with  the  perpendicular. 

3.  From  a  given  point  two  line-segments  are  drawn  to  a  given  straight 
line,  making  unequal  angles  with  the  perpendicular ;  prove  that  the  one 
which  makes  the  greater  angle  with  the  perpendicular  is  the  greater. 

Suggestion.  Let  PA  and  PB  be  the  lines,  drawn  say  on  the  same 
side  of  the  perpendicular  P3I,  PA  making  the  lesser  angle  with  it.  Draw 
PB'  on  the  opposite  side  of  the  perpendicular  making  Z  B'PM-  Z  BP^f. 
Then  Z  PAM  is  greater  than  Z  PBM  and  therefore  greater  than  Z  PB'M. 

4.  Suppose  D  is  any  point  in  one  side  BC  of  a  given  triangle  ABC. 
Find  a  straight  line  such  that  if  the  paper  on  which  the  triangle  is  drawn 
be  folded  along  it,  then  A  will  coincide  with  D. 

5.  A  right  triangle  has  one  of  its  acute  angles  double  the  other.  Prove 
that  the  hypotenuse  is  double  the  lesser  side. 

Suggestion.  If  A  ABC  is  risrht-anscled  at  C,  and  Z  A  double  of  Z  B, 
produce  AC  to  D  making  CD  =  AC.     Prove  ABD  equilateral. 

6.  If  D  is  the  mid-point  ot  BC  the  base  of  an  isosceles  triangle  ABC, 
and  E  any  point  in  AC,  prove  that  the  difference  between  BD  and  DE  is 
less  than  the  difference  between  AB  and  AE. 

7.  On  one  boundary  of  an  angle  whose  vertex  is  A  points  B  and  D 
are  taken,  and  on  the  other  boundary  points  C  and  E,  such  that  AB  is 
equal  to  AC,  and  AD  equal  to  AE ;  prove  that  BE  is  equal  to  CD, 


54 


ELEMENTARY  GEOMETRY 


[Chap.  I 


Proposition  XIX 

78.  If  two  right  triangles  have  the  hypotenuse  and  a 
side  of  one  equal,  respectively,  to  the  hypotenuse  and  a 
side  of  the  other,  the  triangles  are  identically  equal. 


Let  ABC  and  DEF  be  two  triangles,  right-angled  at  C  and 
F,  respectively,  having  the  hypotenuse  AB  equal  to  the  hypot- 
enuse DE,  and  the  side  AC  equal  to  the  side  DF. 

It  is  required  to  prove  that  the  triangles  ABC  and  DEF  are 
identically  equal. 

Proof.  Place  A  ABC  beside  A  DEF  so  that  the  side  AC 
coincides  with  its  equal  side  DF,  the  point  A  with  the  point 
D,  and  the  point  C  with  the  point  F,  and  so  that  the  vertices 
B  and  E  lie  on  opposite  sides  of  DF. 

Then  EF  and  CB  will  lie  in  one  straight  line.     Why  ? 

Since  by  hypothesis  DE  equals  AB,  in  A  DEB,  DE  equals 
DB,  and  therefore  Z  E  equals  Z  B.     Give  reference. 

Again,  if  BC  is  not  equal  to  EF,  then  F  is  not  the  mid-point 
oiEB. 

Suppose  H  is  the  mid-point  of  EB,  and  join  DH. 

Then  DH  must  be  perpendicular  to  EB  (Prop.  VIII),  and 
from  D  there  are  drawn  two  perpendiculars  to  EB,  which  is 
impossible. 

Therefore  H  is  not  the  mid-point  of  EB,  and  similarly  it 
may  be  shown  that  no  other  point  than  F  is  the  mid-point. 

Hence  EF  and  FB  are  equal ;  that  is,  EF  and  BC  are  equal. 

V  Therefore  As  ABC  and  DEF  are  identically  equal. 

\  (Prop.  VIII.) 


78-79]  TRIANGLES  AND  PARALLELOGRAMS  55 


Proposition  XX 

79.  If  two  triangles  have  two  sides  of  the  one  equal, 
respectively,  to  two  sides  of  the  other,  and  the  included 
angles  unequal,  the  triangle  which  has  the  greater 
included  angle  has  also  the  greater  third  side. 


Let  ABC  and  A'B'O  be  the  two  given  triangles  in  which  the 
side  AB  equals  the  side  A'B',  the  side  AC  equals  the  side  A'C, 
and  the  angle  BAC  is  greater  than  the  angle  B'A'C. 

It  is  required  to  prove  that  the  side  BC  is  greater  than  the 
side  B'C. 

Proof.  Place  A  A' B'C  upon  A  ABC  so  that  the  side  A'B' 
will  coincide  with  the  side  AB,  the  point  A'  with  the  point  A, 
and  B'  with  B. 

The  side  A'C  will  not  coincide  with  the  side  AC  [why  ?], 
but  will  fall  within  Z  BAC,  while  the  point  C  may  fall  out- 
side A  ABC  as  in  the  figure,  or  it  may  fall  on  the  side  BC, 
or  within  A  ABC,  according  to  the  lengths  of  the  equal  sides 
AC  and  A'C. 

The  pupil  should  draw  figures  for  the  other  two  cases.     The 
proof  is  the  same  in  all  three  cases. 

Bisect  Z  CAC  by  the  straight  line  AD,  meeting  BC  at  D. 
Join  CD. 

As  AC'D  and  ACD  are  identically  equal.  (Prop.  IV.) 

Therefore  CD  equals  CD. 


66  ELEMENTARY   GEOMETRY  [Chap.  1 

Hence  BD  and  DC  are  togetlier  equal  to  BC. 
But  BD  and  DC  are  togetlier  greater  than  BC.     Why  ? 
Iherelore  BC  is  greater  than  BC.     That  is,  BC  is  greater 
than  B'C. 

Pkoposition  XXI 

80.  If  two  triangles  have  two  sides  of  the  one  equal, 
respectively,  to  two  sides  of  the  other,  and  the  third 
sides  unequal,  the  triangle  which  has  the  greater  third 
side  has  also  the  greater  included  angle. 


Let  ABC  and  A'B'C  be  the  given  triangles  having  the  side 
AB  equal  to  the  side  A'B',  AG  equal  to  A'C,  but  the  third 
side  BC  greater  than  the  third  side  B'C. 

It  is  required  to  prove  that  the  included  angle  BAC  is  greater 
than  the  included  angle  B'A'C. 

Proof.  If  Z  BAC  is  not  greater  than  Z  B'A'C,  it  must  be 
equal  to,  or  less  than  Z  B'A'C. 

First,  suppose  Z  BAC  is  equal  to  Z  B'A'C. 

Then  the  side  BC  must  equal  the  side  B'C  (Prop.  IV.) 

But  this  is  contrary  to  the  hypothesis. 

Next,  suppose  Z  BAC  is  less  than  Z  B'A'C. 

Then  the  side  BC  must  be  less  than  B'C.  (Prop.  XX.) 

But  this  is  also  contrary  to  the  hypothesis. 

Therefore  Z  BAC  must  be  greater  than  Z  B'A'C. 

What  ^ind  of  proof  do  you  call  this  ? 


V9-81J  TRIANGLES  AND   PARALLELOGRAMS  57 

Propositiox  XXII 

81.  //  two  triangles  have  two  sides  of  the  one  equal, 
respectiveli/,  to  ^  two  sides  of  the  other,  and  the  angles 
opposite  one  pair  of  equal  sides  equal,  then  the  angles 
opposite  the  other  pair  of  equal  sides  are  either  equal 
or  supplementary,  and  if  equal,  then  the  triangles  are 
identically  equal. 


tet  ^^^^  and  A'B'C  be  two  triangles  having  the  side  AB 
ec^iial  to  the  side  A'B',  AC  equal  to  A'C,  and  the  angle  at  B 
eqi      ^th*^  angle  at  B',  these  being  opposite  equal  sides. 

It  iffequired  to  prove  that  the  angles  ACB  and  A'C'B'  are 
either  equal  or  supplementary,  and  if  equal,  then  the  tri- 
angles are  identically  equal. 

Proof.  Place  A  A'B'C  upon  A  ABC  so  that  the  side  A'B' 
will  coincide  with  its  equal  side  AB,  the  point  A'  with  the 
point  A,  and  the  point  B'  with  the  point  B. 

Then  since  Z  B'  equals  Z  B,  the  side  B'C  will  fall  on  the 
side  BC,  and  C  will  either  coincide  with  C  or  with  some  other 
point  of  the  straight  line  BC. 

If  C  coincides  with  C,  the  two  triangles  coincide  through- 
out and  are  identically  equal. 

If  C  falls  at  some  other  point,  as  D,  Z  ADC  equals  ZACD 
[why?],  and  Z  ADB  is  supplementary  to  Z  ADC,  or  Z  ACB. 

That  is,  Z  A'C'B'  is  either  equal  to  Z  ACB  or  supplementary 
to  Z  ACB,  and  if  equal  the  triangles  are  identically  equal. 


4 


58  ELEMENTARY  GEOMETRY  [Chap.  I 

82.  Corollary.  If  the  two  angles  given  equal  are  right 
angles  or  obtuse  angles,  the  tivo  triangles  must  be  identically 
equal. 

Proposition  XIX  thus  becomes  a  special  case  of  Proposition  XXII. 

Cases   in  which  Two  Triangles  are  Identically  Equal 

83.  It  may  be  well  at  this  point  to  collect  in  tabulated  form 
the  theorems  which  have  already  been  proved  relating  to  the 
identical  equality  of  two  triangles. 

Two  triangles  are  identically  equal  when  the  following  parts 
in  each  are  respectively  equal : 

1.  Two  sides  and  the  included  angle.  (Prop.  IV.) 

2.  Two  angles  and  the  adjacent  side.  (Prop.  V.) 

3.  The  three  sides.  (Prop.  VIII.) 

4.  A  right  angle,  the  hypotenuse,  and  a  side.  (Prop.  XIX.) 

5.  Two  sides  and  an  angle  opposite  one  of  them.  (Prop.  XXII.) 

It  should  be  noticed  that  in  each  of  the  five  cases  there  are 
three  parts  of  the  one  triangle  equal,  respectively,  to  three 
parts  of  the  other. 

It  must  not,  however,  be  concluded  that  if  any  three  parts 
of  one  triangle  are  respectively  equal  to  the  three  correspond- 
ing parts  of  another,  the  triangles  are  identically  equal.  For 
example,  it  is  not  necessarily  true  that  if  the  three  angles  of 
one  triangle  are  respectively  equal  to  the  three  angles  of 
another,  the  triangles  are  identically  equal. 

In  every  case  it  is  necessary  that  at  least  one  of  the  three 
given  parts  should  be  a  side. 

If  the  three  given  parts  are  two  sides  and  an  angle  opposite 
one  of  them,  as  in  Case  5,  there  is  an  additional  condition  neces- 
sary, namely,  that  the  angles  opposite  one  pair  of  equal  sides 
being  equal,  the  angles  opposite  the  other  pair  shall  also  be 
equal,  not  supplementary. 


82-84]  TRIANGLES  AND   PARALLELOGRAMS  59 


Proposition  XXIII 

84.   At  a  given  point  on  a  given  straight  line  to  con- 
struct an  angle  equal  to  a  given  angle. 


Let  A  be  the  given  angle,  m  the  given  straight  line,  and  0 
the  given  point  on  it. 

It  is  required  to  construct  an  angle  equal  to  A,  having  its 
vertex  at  0,  and  having  the  line  m  for  one  boundary. 

Construction.  With  the  point  xi  as  centre  describe  a  circle 
of  any  radius  cutting  the  boundaries  of  the  given  angle  at  B 
and  C. 

With  0  as  centre  and  the  same  radius  describe  a  circle  cut- 
ting the  line  m  at  the  point  P. 

With  centre  P  and  radius  equal  to  the  line-segment  BC 
describe  a  circle  cutting  the  former  circle  at  Q.     Draw  OQ. 

The  angle  POQ  is  the  required  angle. 

Proof.  Join  BC,  also  PQ.  Prove  A  POQ  identically  equal 
to  ABAC  (Prop.  VIII),  and  hence  Z  0  equal  to  Z  A. 

How  many  such  angles  can  be  constructed  at  0  ? 

EXERCISES 

1.  At  a  given  point  in  a  given  straight  line  construct  an  angle  equal  to 
the  supplement  of  a  given  angle. 

2.  If  the  straight  line  bisecting  the  vertical  angle  of  a  triangle  also 
bisects  the  base,  the  triangle  is  isosceles.     (Apply  Proposition  XXII.) 

8.  Prove  Proposition  VIII  indirectly  by  assuming  the  included  angles 
between  two  pairs  of  corresponding  sides  unequal. 


60  ELEMENTARY   GEOMETRY  [Chap.  I 


MISCELLANEOUS  EXERCISES 

1.  If  one  angle  of  a  triangle  is  equal  to  the  sum  of  the  other  two,  the 
triangle  can  be  divided  into  two  isosceles  triangles. 

2.  If  the  angle  C  of  a  triangle  ABC  is  equal  to  the  sum  of  the  angles 
A  and  B,  the  side  AB  is  equal  to  twice  the  straight  line  joining  C  to  the 
mid-point  of  AB. 

3.  ^  is  a  given  point,  and  B  a  given  point  in  a  given  straight  line.  It 
is  required  to  draw  from  A  to  the  given  straight  line,  a  line-segment  AP 
such  that  the  sum  of  AB  and  BP  may  be  of  given  length. 

4.  If  BE  and  CF  are  drawn  from  the  extremities  of  the  base  of  an 
isosceles  triangle  ABC  meeting  the  equal  sides  at  E  and  F,  and  making 
equal  angles  with  the  bases  then  A  BCF  and  CBE  are  identically  equal. 

5.  The  equal  sides  AB  and  J.C  of  an  isosceles  triangle  are  produced 
to  E  and  F,  respectively,  so  that  AE  equals  AF ;  BF  and  EC  are  joined  ; 
show  (1)  that  5F  equals  EC,  (2)  that  Z  Ci?F  equals  Zi^C^",  (3)  Z  ABC 
equals  /.ACB,  making  use  only  of  Proposition  IV. 

Note.  This  is  Euclid's  order  of  proving  that  the  angles  at  the  base 
of  an  isosceles  triangle  are  equal. 

6.  On  the  same  base  and  on  the  same  side  of  it  there  are  given  two 
triangles,  ACB  and  ADB,  such  that  the  sides  AC  and  AD  are  equal. 
Show  that  the  sides  BC  and  BD  cannot  be  equal. 

Note.     This  is  Euclid's  Proposition  VII  of  Book  I. 

7.  In  a  triangle  ABC  in  which  AB  is  greater  than  AC,  if  the  bisector 
of  the  angle  at  A  meets  BC  a,t  D,  show  that  BB  is  greater  than  CD. 

Suggestion-.  In  AB  make  AE  equal  to  AC,  join  DE.  As  ADE 
and  ADC  are  identically  equal.  Then  apply  Proposition  XII  and  I'ropo- 
sition  XV. 

8.  Show  that  the  line  drawn  from  the  vertex  of  a  triangle  to  any  point 
in  the  base  (between  the  vertices)  is  less  than  the  greater  of  the  two 
sides,  or  than  either  of  them  if  they  are  equal. 

9.  If  D  is  the  mid-point  of  the  side  BC  of  a  triangle  ABC,  in  which 
AC  is  greater  than  AB,  the  angle  ADC  is  obtuse. 

10.  If  in  the  sides  AB,  ^C  of  a  triangle  ABC,  in  which  AC  is  greater 
than  AB,  points  D  and  E  ave  taken  such  that  BD  and  CE  are  equal,  then 
CD  is  greater  than  BE. 


84-S6]  TRIANGLES  AND  PARALLELOGRAMS  61 

Section  II 

PARALLEL    LINES 

85.   If  we  have  given  a  point  0  and  a  straight  line  a,  not 
passing  through  it,  and  draw  through  0  any  straight  line  p. 


unlimited  on  either  side  of  0,  it  will  ordinarily  intersect  a 
in  some  point  A.  If  now  the  line  p  is  rotated  about  0  as  indi- 
cated by  the  arrow-head,  j^  and  a  will  continue  to  intersect  and 
the  point  of  intersection  A  will  move  off  to  the  right.  As 
the  rotation  proceeds,  the  lines  will  by  and  by  intersect  far  out 
to  the  left,  but  the  point  of  intersection  will  still  move  from 
the  left  to  the  right. 

The  question  arises.  Do  these  lines  always  intersect?  and 
we  are  accustomed  to  say,  No,  there  is  one  position  of  p  for 
which  they  do  not  intersect.  When  p  is  in  that  position  it  is 
said  to  ba  parallel  to  a. 

Defixition.  Parallel  straight  lines  are  straight  lines  lying 
in  the  same  plane,  and  such  that  if  they  are  drawn  ever 
so  far  either  way  they  will  not  intersect. 

Straight  lines  in  the  same  plane  which  are  not  parallel  must 
intersect,  if  produced  far  enough. 

86.  Postulate  5.  Through  ariy  jwint  one  and  only  one 
straight  line  can  he  drawn  parallel  to  a  given  straight  line. 


62 


ELEMENTARY  GEOMETRY 


[Chap.  I 


87.  If  two  straight  lines,  parallel  or  not,  are  intersected  by 
a  third  line,  called  a  transversal,  four  angles  are  made  at  each 
intersection.  These  may  be 
designated  by  writing  a  let- 
ter in  each  angle,  and  calling 
the  angle  by  the  name  of  the 
letter. 

At  each  intersection  there 
are  two  pairs  of  vertical  angles 
and  four  pairs  of  adjacent  an- 
gles. 

Other  pairs  of  angles  have 
also  received  special  names. 
6',  c  and  c',  d  and  d'  are  called  corresponding  angles. 

The  angles  a  and  d',  b  and  c',  c  and  b',  d  and  a',  are  called 
alternate  angles. 

The  angles  a,  b,  c',  d',  are  called  exterior  angles,  and  the 
angles  c,  d,  a',  b',  are  called  interior  angles. 


Thus  the  angles  a  and  a',  b  and 


Proposition  XXIV 

88.  //  two  straight  lines  lying  in  one  plane  are  such 
that  a  transversal  makes  the  interior  alternate  angles 
equal  these  two  lines  are  parallel. 


Let  the  straight  lines  a  and  b  lie  in  one  plane  and  be  such 
that  the  transversal  c  makes  the  alternate  angles  m  and  n  equal. 


87-92]  TRIANGLES  AND  PARALLELOGRAMS  68 

It  is  required  to  prove  that  a  and  b  are  parallel. 

Proof.  If  a  and  b  are  not  parallel  they  will  intersect  if 
produced  far  enough  either  one  way  or  the  other. 

If  they  intersect,  they  and  the  transversal  will  form  a  tri- 
angle in  which  one  of  the  two  angles  m  and  n  will  be  an  exterior 
angle  formed  by  producing  a  side,  and  the  other,  an  interior 
non-adjacent  angle,  in  which  case  they  could  not  be  equal. 

(Prop.  XII.) 

Hence  a  and  b  cannot  intersect,  and  therefore  must  be 
parallel. 

Exercise.  Show  that  if  any  one  pair  of  alternate  angles  are  equal, 
all  pairs  are  equal. 

89.  Corollary  I.  If  two  straight  lines  lying  in  one  plane 
are  such  that  a  transversal  makes  a  pair  of  corresponding  angles 
equal,  these  two  lines  are  parallel. 

Prove  by  showing  that  if  a  pair  of  corresponding  angles  are 
equal  a  pair  of  alternate  angles  must  also  be  equal. 

90.  Corollary  II.  If  two  straight  lines  lying  in  one  plane 
are  such  that  a  transversal  makes  a  2')air  of  interior  angles,  or  a 
pair  of  exterior  angles,  on  the  same  side  of  it  supplementary, 
these  two  lines  are  parallel. 

91.  Corollary  III.  Straight  lines  lying  in  one  plane  and 
perpendicular  to  the  same  straight  line  are  parallel. 

92.  Corollary  IV.  Two  straight  lines  parallel  to  the  same 
straight  line  are  parallel  to  each  other. 

For,  if  they  are  not  parallel  they  must  intersect  when  pro- 
duced ;  then  there  would  be  through  their  point  of  inter- 
section two  straight  lines  parallel  to  the  same  line, 
contrary  to  Postulate  5. 


64  ELEMENTARY  GEOMETRY  [Chap.  I 


Proposition  XXV 

93.   If  two  parallel  straight  lines  are  cut  by  any  trans- 
versal, the  interior  alternate  angles  are  equal. 


Let  a  and  h  be  two  parallel  lines  and  c  any  transversal  inter- 
secting a  at  J.  and  h  at  B. 

It  is  required  to  prove  that  the  interior  alternate  angles  m  and 
n  are  equal. 

Proof.  If  Z  m  is  not  equal  to  Z  n,  through  A  draw  the 
straight  line  a'  making  with  the  transversal  c  an  angle  m^ 
which  is  equal  to  Z  n. 

Then  the  straight  lines  a'  and  b  must  be  parallel.  (Prop. 
XXIV.) 

Bat,  by  hypothesis,  a  and  h  are  parallel. 

Therefore,  since  through  A  but  one  straight  line  can  be  drawn 
parallel  to  h,  a'  must  coincide  with  a,  and  Z  ?/i'  must  coincide 
with  Z  771. 

But  by  construction  Zm'  =  Z.  n. 

Therefore  Zm  =  Z  n. 

94.  Corollary  I.  If  two  parallel  lines  are  cut  hi/  a  trans- 
versal, any  two  alternate  angles  and  any  two  corresponding  angles 
are  equal. 


93-98]  TRIANGLES  AND   PARALLELOGRAMS  ^b 

95.  Corollary  II.  If  two  parallel  lines  are  cut  by  any  trans- 
versal, the  interior  angles,  and  also  the  exterior  aiigles,  on  the 
same  side  of  the  transversal,  are  supplementary. 

96.  Corollary  III.  If  any  pair  of  alterjiate  angles  foryned 
by  two  straight  lines  and  a  transversal  are  unequal,  or  if  any 
pair  of  corresponding  angles  are  unequal,  or  if  the  interior  angles 
on  one  side  are  7wt  supplementary,  the  two  lines  are  not  parallel 
and  therefore  will  meet  if  produced. 

For  if  the  lines  are  parallel,  we  have  just  proved  that  alternate 
angles  must  be  equal,  corresponding  angles  must  be  equal, 
and  interior  angles  on  one  side  must  be  supplementary,  con- 
trary to  the  assumption. 

97.  Corollary  IV.  If  a  straight  lirie  is  perpendicular  to 
one  of  two  parallel  lines,  it  is  perpendicular  also  to  the  other. 

98.  Corollary  V.  If  two  straight  lines  intersect,  lines  per- 
pendicular to  them  in  the  same  plane  will  also  intersect. 

Let  a  and  b  be  two  intersecting  straight 
lines  and  let  c  be  perpendicular  to  a,  and  d 
perpendicular  to  b.  If  c  and  d  do  not  in- 
tersect, they  must  be  parallel,  and  then 
since  a  is  perpendicular  to  c  it  must  also 
be  perpendicular  to  d  (Cor.  IV).  Hence 
a  and  b  are  both  perpendicular  to  d  and 
must  themselves  be  parallel  (Art.  91), 
contrary  to  hypothesis. 

EXERCISES 

1.  Any  straight  line  drawn  parallel  to  the  base  of  an  isosceles  triangle 
forms  with  the  sides  another  isosceles  triangle. 

2.  Is  it  always  true  that  if  two  angles  are  equal  and  one  boundary  of 
the  one  parallel  to  one  boundary  of  the  other,  then  their  other  boundaries 
are  also  parallel  ? 

Illustrate  your  answer  by  diagrams. 

3.  Prove  that  two  parallel  lines  are  everywhere  equidistant. 

4.  If  any  two  straight  lines  intersect,  lines  parallel  to  them  must  also 
intersect. 


66  ELEMENTARY  GEOMETRY  [Chap.  1 


Proposition  XXVI 

99.   Through  a  given  point  to  draw  the  straight  line 
parallel  to  a  given  straight  line. 


v^  J^ 


X 


Let  A  be  the  given  point  and  BC  the  given  straight  line. 

It  is  required  to  draw  through  A  the  straight  line  parallel  to 
BC. 

Construction.  Through  A  draw  any  straight  line  intersect- 
ing BC  at  a  point  D. 

At  A,  in  the  straight  line  AD,  construct  Z  DAE  equal  to 
AADB.  (Prop.  XXIII.) 

Then  the  straight  line  AE  is  parallel  to  BC.     (Prop.  XXIV.) 

EXERCISES 

1.  Through  a  given  point  draw  a  straight  line  making  with  a  given 
straight  Hne  an  angle  equal  to  a  given  angle.  How  many  such  lines  can 
be  drawn  ? 

2.  Two  line-segments  AB  and  CD  bisect  each  other  at  0.  Prove  that 
the  lines  AC  and  BD  are  parallel.     How  about  AD  and  BC  ? 

3.  If  through  the  vertex  of  an  isosceles  triangle  a  straight  line  is  drawn 
parallel  to  the  base,  it  will  bisect  the  exterior  vertical  angle. 

4.  Through  a  given  point  draw  a  straight  line  which  will  make  equal 
angles  with  two  given  intersecting  straight  lines. 

SuGGESTioy.  Draw  the  perpendicular  to  the  bisector  of  the  angle  be- 
tween the  lines. 


99-103]         TRIANGLES  AND  PARALLELOGRAMS  67 

Proposition  XXVII 

100.  //  one  side  of  a  triangle  is  produced,  the  exterior 
angle  so  formed  is  equal  to  the  sum  of  the  two  interior 
hon-adjacent  angles. 


Let  ABC  be  any  triangle  having  the  side  BC  produced  to  D. 

It  is  required  to  prove  that  the  exterior  angle  ACD  is  equal 
to  the  sum  of  the  two  interior  non-adjacent  angles,  viz.  the 
angles  CAB  and  CBA. 

Proof.     From  C  draw  the  straight  line  CE  parallel  to  BA. 

(Prop.  XXVI.) 

Then  Z  BCD  =  Z  CBA  and  Z  ACE  =  Z  CAB.     Why  ? 

Therefore  the  sum  of  Zs  ACE  and  ECD  equals  the  sum  of 
Zs  CAB  and  CBA 

That  is,  /-ACD  equals  the  sum  of  Zs  CAB  and  CBA. 

101.  Corollary  I.  Tlie  sum  of  the  three  angles  of  a  triangle 
is  equal  to  two  right  angles. 

102.  Corollary  II.  The  two  acute  angles  of  any  right  tri- 
angle are  complementary ;  i.e.  their  sum  is  a  right  angle. 

Definition.  One  angle  is  said  to  be  the  complement  of 
another  when  the  sum  of  the  two  is  a  right  angle. 

103.  Corollary  III.  Each  angle  of  a  triangle  is  the  supple- 
ment of  the  sum  of  the  other  two. 


68  ELEMENTARY  GEOMETRY  [Chap.  I 

104.  Corollary  IV.  If  tico  triangles  have  tico  angles  of  the 
one  equal  respectively  to  tico  angles  of  the  other ^  their  third  angles 
are  also  equal;  or,  if  the  sum  of  two  angles  in  the  one  is  equal  to 
the  sum  of  two  angles  in  the  other,  their  third  angles  are  equal. 

• 

105.  Corollary  V.  If  the  boundaries  of  one  angle  are  re- 
spectively parallel  or  perpendicidar  to  the  boundaries  of  another, 
these  two  angles  are  either  equal  or  supplementary. 

Make  diagrams  representing  all  the  cases. 

106.  Corollary  VI.  If  two  triangles  have  two  angles  of  the 
one  equal  respectively  to  two  aiigles  of  the  other,  and  the  sides 
opposite  one  pair  of  equal  angles  also  equal,  the  triangles  are 
identically  equal. 

For  then  the  third  angles  are  also  equal,  and  the  theorem  falls 
under  Proposition  V.  This  theorem  may  be  easily  proved 
by  superposition,  making  use  of  Proposition  XII. 


EXERCISES 

1.  If  an  isosceles  triangle  is  right-angled,  each  of  the  angles,  at  the 
base  is  half  a  right  angle, 

2.  If  two  isosceles  triangles  have  their  vertical  angles  equal,  they  are 
mutually  equiangular  ;  i.e.  each  angle  of  the  one  is  equal  to  an  angle  of 
the  other. 

3.  If  one  angle  of  a  triangle  is  equal  to 
the  sum  of  the  other  two,  it  must  be  a  right 
angle. 

4.  Divide  a  right  triangle  into  two  isos- 
celes triangles,  and  hence  show  that  the  mid- 
point of  the  hypotenuse  is  equidistant  from 
the  three  vertices. 

Suggestion.  Construct  /.ACT)  =  ZA,  and  show  that  Z BCD  =  ZB. 
Hence  DA  =  DC  ^  DB. 

6.    Each  angle  of  an  equilateral  triangle  is  two-thirds  of  a  right  angle. 


104-107]        TRIANGLES  AND  PARALLELOGRAMS  69 

Proposition  XXVIII 

107.  Every  point  on  the  bisector  of  an  angle  is  equi- 
distant from  the  boundaries  of  the  angle;  and  every 
point  within  the  angle  which  is  equidistant  from 
the  boundaries  is  on  the  bisector;  that  is,  the  bisector 
of  an  angle  is  the  locus  of  points  within  the  angle 
which  are  equidistant  from  its  boundaries. 


First,  let  AD  be  the  bisector  of  the  angle  BAG  and  let  P 
be  any  point  on  AD. 

It  is  required  to  prove  that  P  is  equidistant  from  AB 
and  AC. 

Proof.  From  P  draw  the  lines  PM  and  PN  peri^endicular, 
respectively,  to  AB  and  AC. 

Then  As  PAM  and  PAN  are  identically  equal.  Why? 
(Apply  Prop.  V.) 

Therefore  PM  equals  PN-,  that  is,  P  is  equidistant  from 
AB  and  AC. 

Next,  let  Q  be  any  point  equidistant  from  AB  and  AC 
within  the  angle  BAC. 

It  is  required  to  prove  that  Q  lies  on  the  bisector  of  the  angle 
BAC. 

Proof.  From  Q  draw  QS  and  QT  perpendicular  to  AB  and 
AC,  respectively,  and  join  QA. 

By  hypothesis,  QS  and  QTare  equal. 


70  ELEMENTARY  GEOMETRY  [Chap.  I 

Then  As  QSA  and  QTA  are  identically  equal.  Why? 
(Apply  Prop.  XIX.) 

Therefore  /LQAS  equals  ZQAT;  that  is,  Q  lies  on  the 
bisector  of  the  angle  BAC. 

Since  the  bisector  of  the  angle  BAC  contains  all  points 
within  the  angle  which  are  equidistant  from  the  boundaries, 
and  no  point  which  is  not  equidistant  from  the  boundaries,  it 
is  the  locus  of  such  points. 

108.  Corollary.  The  locus  of  jwlnts  equidistant  from  two 
intersecting  straight  lines  consists  of  the  two  lines  which  bisect 
the  angles  formed  by  the  given  lines. 

EXERCISES 

1.  Prove  that  the  locus  of  points  equidistant  from  two  intersecting 
straight  lines  consists  of  two  straight  lines  at  right  angles. 

2.  If  the  straight  line  drawn  from  one  vertex  of  a  triangle  to  the  mid- 
point of  the  opposite  side  is  equal  to  half  of  this  side,  prove  that  the 
triangle  has  one  right  angle. 

3.  If,  in  a  right  triangle,  a  perpendicular  is  drawn  from  the  vertex 
of  the  right  angle  to  the  opposite  side,  the  two  triangles  so  formed  are 
equiangular  with  each  other  and  with  the  whole  triangle. 

4.  A  straight  line  drawn  perpendicular  to  the  base  5  C  of  an  isosceles 
triangle  ABC  meets  the  side  AB  at  E  and  the  side  CA  produced  at  F. 
Prove  that  the  triangle  EAF  is  isosceles. 

6.  If  a  point  is  equidistant  from  two  parallel  straight  lines,  any  line- 
segment  drawn  through  it  and  terminated  by  the  parallel  lines  is  bisected 
at  the  point. 

6.  If  a  point  is  equidistant  from  two  parallel  straight  lines,  any  two 
straight  lines  drawn  through  it  intercept  equal  segments  of  the  parallel 
lines. 

7.  Construct  a  triangle  having  given  two  angles  and  the  length  of  the 
perpendicular  from  the  third  vertex  to  the  opposite  side. 

8.  If  two  exterior  angles  of  a  triangle  be  bisected,  and  from  the  point 
of  intersection  of  the  bisecting  lines  a  straight  line  is  drawn  to  the  third 
vertex,  it  bisects  the  third  angle.     (Apply  Proposition  XXVIII.) 


107-109]        TRIANGLES  AND  PARALLELOGRAMS 


71 


Section  III 

CLOSED  RECTILINEAR   FIGURES   OF   MORE   THAN 
THREE   SIDES 


109.  On  page  13  a  closed  figure  was  defined  as  one  which 
can  be  traversed  by  starting  at  any  point  of  it,  and  moving 
continuously  along  the  lines  of  the  figure  in  order,  returning 
to  the  same  point  without  passing  twice  over  any  portion  of 
the  figure. 

A  closed  rectilinear  figure  is  one  which  is  made  up  wholly  of 
line-segments  and  the  points  in  which  they  intersect,  two  and 
two,  in  order. 


The  line-segments  are  called  the  sides  of  the  figure,  and  the 
points  in  which  the  sides  intersect  are  called  the  vertices  of 
the  figure. 

Two  sides  which  intersect  in  a  vertex  are  called  adjacent 
sides. 

The  angles  formed  by  pairs  of  adjacent  sides  are  called  the 
angles  of  the  figure. 

A  straight  line  joining  any  two  vertices  not 
on  the  same  side  is  a  diagonal. 

In  the  diagram,  the  points  A,  B,  C,  D,  E 
are  vertices,  the  line-segments  AB,  BC,  CD, 
etc.,  are  sides,  and  AC,  AD,  CE,  etc.,  are 
diagonals. 

If  the  number  of  sides  of  any  closed  rectilinear  figure  is  n, 
the  number  of  diagonals  which  can  pass  through  any  one 
vertex  is  n  — 3,  and  since  each  diagonal  passes  through  two 
vertices,  the  total  number  of  diagonals  is  ^n{n  —  3). 


72  ELEMENTARY  GEOMETRY  [Chap.  1 

110.   A  convex  rectilinear  figure  is  one  which  lies  wholly  on 
one  side  or  the  other  of  each  of  its  sides. 


Figures  1  and  2  above  are  convex  rectilinear  figures,  while 
figures  3  and  4  are  closed  but  not  convex. 

Could  a  triangle  be  anything  else  than  convex  ? 

111.  We  shall  usually  denote  a  closed  rectilinear  figure  by  placing 
capital  letters  at  its  vertices  and  naming  them  in 

order.  Thus,  the  figure  at  the  right  is  the  figure 
ABCDEF,  or  CBAFED,  etc.  When  no  misunder- 
standing is  likely  to  arise,  the  figure  may  be  named 
by  mentioning  two  non-consecutive  vertices. 

Sometimes  it  will  be  convenient  to  letter  the  sides 
of  the  figure  a,  6,  c,  d,  e,f,  and  then  the  figure  can 
be  named  by  mentioning  the  sides  in  order. 

112.  It  will  be  observed  that  in  any  closed  rectilinear  figure  there  are 
just  as  many  vertices  as  sides.  The  vertices  and  sides  together  make  up 
the  elements  of  the  figure,  so  that  the  number  of  elements  is  always  even, 
and  equal  to  twice  the  number  of  sides  of  the  polygon. 

113.  Two  elements  are  said  to  be  opposite  when  just  as  many  ele- 
ments lie  between  them  the  one  way  round  as  the  other  way  round  the 
figure. 

If  the  number  of  sides  in  the  figure  is  even,  a  side  is  always  opposite  a 
side,  and  a  vertex  opposite  a  vertex.  lu  the  figure  above,  the  side  CD  is 
opposite  the  side  AF,  the  vertex  B  opposite  the  vertex  E,  etc.  the  number 
of  elements  between  the  pairs  of  opposite  elements  being  in  each  case 
five. 

If  the  number  of  sides  in  the  figure  is  odd,  a  side  is  always  opposite 
a  vertex. 

We  have  already  assumed  that  the  pupil  will  know  the  opposite  ele- 
ments in  a  triangle  ;  it  is  not,  however,  so  easy  to  detect  opposite  elements 
when  the  figure  has  more  than  three  sides. 


110-llG]        TRIANGLES  AND  PARALLELOGRAMS  78 

114.  A  closed  rectilinear  figure  having  three  sides  is  a  tri- 
angle, one  of  four  sides  is  a  quadrilateral. 

A  general  name  applied  to  any  closed  rectilinear  figure,  with- 
out regard  to  the  number  of  sides,  is  a  polygon. 

115.  A  convex  polygon  is  a  closed  rectilinear  figure  which 
lies  wholly  on  one  side  or  the  other  of  each  of  its  sides. 

Proposition  XXIX 

116.  The  sum  of  the  interior  angles  of  any  convex 
polygon,  together  with  four  right  angles,  uxahes  up 
twice  as  many  right  angles  as  the  polygon  has  sides. 


Let  ABODE  be  any  convex  polygon. 

It  is  required  to  prove  that  the  sum  of  the  interior  angles 
A,  B,  C,  D,  E,  together  with  four  right  angles,  makes  up  twice 
as  many  right  angles  as  the  figure  has  sides. 

Proof.  Choose  any  point  O  within  the  polygon,  and  join 
0  to  each  vertex  of  the  polygon. 

This  divides  the  polygon  into  as  many  triangles  as  the  poly- 
gon has  sides. 

The  three  angles  of  each  triangle  are  together  equal  to  two 
right  angles.  (Prop.  XXVII,  Cor.  I.) 

Hence  all  the  angles  of  all  the  triangles  are  together  equal 
to  two  right  angles  for  each  side,  or  to  twice  as  many  right 
angles  as  the  polygon  has  sides. 

But  these  angles  include  the  interior  angles  of  the  polygon, 
together  with  four  right  angles  about  the  point  0. 

Therefore  .  .  . 


74  ELEMENTARY  GEOMETRY  [Chap.  I 

117.  Corollary  I.     If  the  sides  of  a  con-  ^^ 

vex  polygon  are  produced  in  order,  the  exterior  >^\^  / 

angles  so  formed  are  together  equal  to  four  right      ,*'\  j 

angles.  \  / 

Suggestion.     Each  exterior  angle  and  its  adjacent  \ 

interior  angle  make  two  right  angles.  \ 

118.  Corollary  II.  If  S  represents  the  sum  of  the  interior 
angles  of  a  convex  polygon,  and  n  the  number  of  sides,  then  S  4- 
four  right  angles  =  2n  right  angles,  or  S  =  (n  —  2)  times  two 
right  angles. 

Definitions 

119.  A  parallelogram  is  a  quadrilateral  whose  opposite  sides 
are  parallel. 

If  only  two  opposite  sides  of  a  quadrilateral  are  parallel, 
the  figure  is  called  a  trapezoid. 

If  no  two  sides  of  a  quadrilateral  are  parallel,  the  figure  is 
called  a  trapezium. 

120.  A  parallelogram  having  two  adjacent  sides  equal  is 
called  a  rhombus. 

121.  A  parallelogram  having  one  right  angle  is  called  a 
rectangle. 

122.  A  square  is  a  parallelogram  having  one  right  angle 
and  two  adjacent  sides  equal. 

EXERCISES 

1.  The  sum  of  the  interior  angles  of  any  quadrilateral  is  equal  to  four 
right  angles. 

To  how  many  right  angles  is  the  sum  of  the  interior  angles  of  a  con- 
vex polygon  of  five  sides  equal  ?  of  six  sides  ?  of  eight  sides  ? 

2.  If  two  angles  of  a  quadrilateral  are  right  angles,  the  other  two  are 
supplementary. 

3.  If  any  one  angle  of  an  isoceles  triangle  equals  two  thirds  of  a  right 
angle,  the  triangle  is  equilateral. 


117-125]       TRIANGLES  AND  PARALLELOGRAMS  75 


Proposition  XXX 

123.   The  opposite  sides  and  the  opposite  angles  of  a 
parallelogram  are  equal. 


D 

Let  ABCD  be  a  parallelogram 

It  is  required  to  prove  that  AB  is  equal  to  DC  and  AD  equal 
to  BC\  also  that  the  angle  A  is  equal  to  the  angle  C,  and  the 
angle  B  equal  to  the  angle  D. 

Proof.     Join  a  pair  of  opposite  vertices  A  and  C. 

Because  AB  is  jjarallel  to  DC,  by  definition, 

Z  BAC  =  Z  DC  A.  (Prop.  XXV.) 

And  because  AD  is  parallel  to  BC, 

ZBCA^ZDAG. 

Therefore  As  ABC  and  CDA  are  identically  equal.    (Prop.  V.) 

Conclusion  .  .  . 

124.  Corollary.  Either  diagonal  of  a  x>arallelogram  divides 
the  figure  into  two  superpo sable  triangles. 

125.  In  writing  out  the  proofs  of  propositions  and  in  the 
exercises,  the  pupil  is  advised  to  make  use  of  certain  well 
established  symbols  in  order  to  abbreviate  his  work.  For 
instance,  instead  of  the  expression  *is  parallel  to,'  or  the  word 
*^ parallel,'  he  may  use  the  symbol  ||  ;  and  instead  of  4s  perpen- 
dicular to,'  or  ^  perpendicular,'  he  may  use  the  symbol  X.  Thus 
'AB  is  parallel  to  CD'  would  be  written  'AB  ||  CD;  and  '  AB 
is  perpendicular  to  OZ)'  would  be  written  'AB±CD\  The 
pupil  is  also  advised  to  make  free  use  of  algebraic  symbols  in 
all  written  work. 


76  ELEMENTARY   GEOMETRY  [Chap. 


Proposition    XXXI 

126.   The  diagonals   of  a  parallelogram  bisect  each 
other. 

A B 


Let  ABCD  be  a  parallelogram  whose  diagonals  intersect  at  0. 

It  is  required  to  prove  that  0  is  the  mid-point  of  both  AC 
and  BD. 

Proof.  Compare  the  two  triangles  AGE  and  COD,  making 
use  of  Propositions  XXV,  XXX,  and  V. 

Proposition   XXXII 

127.  //  two  sides  of  a  quadrilateral  are  parallel  and 
equal,  the  figure  is  a  parallelogram. 


Let  ABCD  be  a  quadrilateral  having  the  sides  AB  and  Z)C 
parallel  and  equal. 

It  is  required  to  prove  that  AD  and  BC  are  parallel,  and 
hence  that  the  figure  is  a  parallelogram. 

Proof.     Draw  a  diagonal  AC.     Then  As  BAC  and  DCA  are 
identically  equal  by  Propositions  XXV  and  IV. 

Therefore  Z  BCA  =  Z  DAC, 

and  the  sides  AD  and  BC  are  parallel.  (Prop.  XXIV.) 


126-129]        TRIANGLES  AND  PARALLELOGRAMS  77 

Proposition   XXXIII 

128.  If  both  pairs  of  opposite  sides  of  a  quadrilateral 
are  equal,  the  figure  is  a  parallelogram. 

Given  a  quadrilateral  whose  opposite  sides  are  equal. 
It  is  required  to  prove  that  the  opposite  sides  are  parallel, 
and  hence  that  the  figure  is  a  parallelogram. 

Proof.    Draw  a  diagonal  and  compare  the  triangles  so  formed. 
[The  pupil  should  make  the  necessaiy  diagram  for  himself.] 

Proposition  XXXIV 

129.  //  two  parallelograms  have  two  adjacent  sides 
and  the  included  angle  of  the  one  equal,  respectively,  to 
two  adjacent  sides  and  the  included  angle  of  the  other, 
the  parallelograms  are  identically  equal. 

Prove  by  superposition,  placing  the  one  parallelogram  on  the 
other  so  that  parts  given  equal  will  coincide,  and  showing  that 
the  remaining  parts  must  also  coincide. 

EXERCISES 

1.  If  the  diagonals  of  a  quadrilateral  bisect  each  other,  the  quadrilateral 
is  a  parallelogram. 

2.  In  a  parallelogram  the  perpendiculars  drawn  from  one  pair  of  oppo- 
.  site  vertices,  to  the  diagonal  through  the  other  pair,  are  equal. 

3.  If  the  diagonals  of  a  parallelogram  are  equal,  the  parallelogram  is 
rectangular. 

4.  If  ABCD  is  a  parallelogram,  and  X,  F,  respectively,  the  mid-points 
of  the  sides  AD,  BC ;  show  that  the  figure  ^  FOX  is  a  parallelogram. 

5.  If  both  pairs  of  opposite  angles  of  a  quadrilateral  are  equal,  the 
figure  is  a  parallelogram. 

6.  If  one  angle  of  a  parallelogram  is  a  right  angle,  all  the  angles  are 
right  angles. 

7.  If  two  sides  of  a  quadrilateral  are  parallel,  and  the  other  two  equal 
but  not  parallel,  the  diagonals  are  equal. 

Such  a  figure  is  called  an  isosceles  trapezoid. 


78  ELEMENTARY  GEOMETRY  [Chap.1 

Section  IV 

MISCELLANEOUS  THEOREMS 

Proposition  XXXV 

130.  T?ie  line- segment  joining  the  mid-points  of  two 
sides  of  a  triangle  is  parallel  to  the  third  side  and 
equal  to  half  of  it. 

A 


Let  ABC  be  any  triangle,  D  and  E  the  mid-points  of  the  two 
sides  AB  and  AC. 

It  is  required  to  prove  that  the  line-segment  DE  is  parallel  to 
the  third  side  BC,  and  equal  to  half  of  it. 

Proof.  From  B  draw  5ii^  parallel  to  CA,  meeting  the  straight 
line  ED  in  F. 

A  BDF  is  identically  equal  to  A  ADE.     Prove. 

Therefore  DF  =  DE,  and  BF=AE. 

But  AE  =  CE  by  hypothesis. 

Therefore  BF  =  CE,  and  is  parallel  to  CE  by  construction. 

Therefore  BFEC  is  a  parallelogram  (Prop.  XXXII),  and  FE, 
i.e.  DE,  is  parallel  to  BC. 

It  is  left  for  the  pupil  to  prove  that  DE  equals  half  of  BC. 

131.  Corollary  I.  If  the  mid-points  of  the  sides  of  any 
quadrilateral  be  joined  in  order,  the  figure  so  formed  is  a  paral- 


130-135]        TRIANGLES  AND  PARALLELOGRAMS  79 

lelogranij  and  the  sum  of  the  sides  of  this  parallelogram  equals 
the  sum  of  the  diagonals  of  the  quadrilateral. 

Suggestion.  Draw  the  diagonals  of  the  quadrilateral  and  apply 
Proposition  XXXV. 

132.  Definition.  The  sum  of  the  sides  of  any  polygon  is 
called  the  perimeter  of  the  polygon. 

133.  Corollary  II.  If  from  the  mid-point  of  one  side  of  a 
triangle  there  is  drawn  a  straight  line  parallel  to  a  second  side^ 
this  line  passes  through  the  mid-point  of  the  third  side. 

Else  through  one  point  there  could  be  drawn  two  parallels  to 
the  same  line. 

134.  Definition.  A  straight  line  drawn  from  any  vertex 
of  a  triangle  to  the  mid-point  of  the  opposite  side  is  called 
a  median  of  the  triangle.     Every  triangle  has  three  medians. 

135.  Definition.  By  concurrent  lines  we  mean  lines  which 
pass  through  the  same  point. 

EXERCISES 

1.  The  mid-points  of  a  pair  of  opposite  sides  of  a  quadrilateral  and  the 
mid-points  of  the  diagonals  are  the  vertices  of  a  parallelogram. 

2.  In  a  trapezoid  the  mid-points  of  the  non-parallel  sides  and  the  mid- 
points of  the  diagonals  lie  upon  the  same  straight  line  parallel  to  the  other 
two  sides.  The  distance  between  the  mid-points  of  the  non-parallel  sides 
is  equal  to  half  the  sum  of  the  parallel  sides  ;  and  the  distance  between 
the  mid-points  of  the  diagonals  is  equal  to  half  the  difference  between  the 
parallel  sides. 

3.  Find  on  one  side  of  a  triangle  the  point  from  which  straight  lines 
drawn  parallel  to  the  other  two  sides,  and  terminated  by  these  sides,  are 
equal. 

Suggestion,     Bisect  the  opposite  angle. 

4.  It  ABCD  and  AEFG  are  two  squares  so  situated  that  the  angles  A 
coincide,  then  A,  C,  and  F  are  in  the  same  straight  line. 

6.  If  two  parallelograms  have  a  diagonal  and  two  vertices  in  common, 
show  that  the  other  four  vertices  are  the  vertices  of  another  parallelogram. 


80 


ELEMENTARY  GEOMETRY 


[Chaf.  1 


Proposition  XXXVI 
136.   The  medians  of  a  triangle  are  concurrent. 

B 


Let  ABC  be  any  triangle,  AE  and  CD,  two  of  its  medians 
meeting  at  0.     Join  BO  and  produce  to  meet  AC  at  F. 

It  is  required  to  prove  that  F  is  the  mid-point  of  AC,  and 
hence  that  BF  is  the  third  median  of  the  triangle. 

Proof.     Through  A  draw  AK  parallel  to  DC,  meeting  BF 
at  K.     Join  CK. 

Then  in  A  BAK,  since  BD  =  DA,  BO  =  OK. 

(Prop.  XXXV.,  Cor.  II.) 

In  ABCK,  since  BE  =  EC,  and  BO  =  OK, 

EO  is  parallel  to  CK;   that  is,  OA  is  parallel  to  CK 

Therefore  AOCK  is  a  parallelogram,  and  AF  equals  FC 
Therefore  BF,  passing  through  0,  is  a  median  of  the  triangle. 

(Prop.  XXXI.) 

137.  Definition.    The  point  0,  in  which  the  medians  inter- 
sect, is  called  the  centroid  of  the  triangle. 

138.  Corollary.     The  centroid  of  a  triangle  is  a  trisection 
point  of  each  median  of  the  triangle. 

For,  since  0K=  OB,  and  0F=\  OK,  therefore,  OF=l  OB, 

or  \  BF. 
Similarly  O  may  be  shown  to  be  a  point  of  trisection  of  the 

other  medians. 


136-139]        TRIANGLES  AND  PARALLELOGRAMS  81 


Proposition   XXXVII 

139.   The  bisectors  of  the  angles  of  a  triangle  are  con- 
current. 

A 


Let  ABC  be  any  triangle,  and  AO^,  BO2,  CO^,  the  bisectors 
of  the  angles. 

It  is  required  to  prove  that  AOi^  BO2,  and  CO3  meet  in 
one  point,  or  are  concurrent. 

Proof.  Since  Zs  ABO2  and  BAOi  are  together  less  than 
two  right  angles  [why?],  AOi  and  BO^  will  meet. 

(  Prop.  XXV,  Cor.  III.) 

Let  them  meet  in  X. 

Now  AOi  is  the  locus  of  points  within  the  angle  BAC 
equidistant  from  AB  and  AC.  (Prop.  XXVIII.) 

Also,  BO2  is  the  locus  of  points  within  the  angle  ABC 
equidistant  from  AB  and  AC. 

The  point  X  must  therefore  be  equidistant  from  AC  and  BC, 
and  consequently  must  lie  on  CO3,  which  is  the  locus  of  points 
within  the  angle  ACB  equidistant  from  AC  and  BC. 

Therefore  AOi,  BO2,  CO^  meet  in  the  point  X 

1.  Produce  the  sides  of  the  triangle  and  bisect  two  external  angles  and 
the  third  internal  angle.     Are  these  bisectors  concurrent  ?     Prove. 

By  this  means  we  find  four  points  X,  Y,  Z,  W  associated  with  the  tri- 
angle. Notice  that  when  the  figure  is  completed,  these  four  points  are 
joined  in  all  possible  ways,  and  that  the  line  joining  any  two  of  them  is 
perpendicular  to  the  line  joining  the  other  two. 


82  ELEMENTARY  GEOMETRY  [Chap.  I 


Proposition   XXXVIII 

140.  The  perpendicular  bisectors  of  the  sides  of  a 
triangle  are  concurrent. 

Let  ABC  be  any  triangle ;  i),  E^  F,  the  mid-points  of  the 
sides. 

It  is  required  to  prove  that  the  perpendiculars  erected  at 
D,  E,  and  F  meet  in  a  point. 

Proof.  The  perpendicular  at  F  is  the  locus  of  points  equi- 
distant from  A  and  B.  (Art.  72.) 

The  perpendicular  at  E  is  the  locus  of  points  equidistant 
from  A  and  C. 

These  two  perpendiculars  intersect.     Why? 

Their  point  of  intersection  being  equidistant  from  B  and 
(7,  must  lie  on  the  perpendicular  at  D.  (Art.  74.) 

Therefore  the  three  perpendiculars  meet  in  a  point. 

EXERCISES 

1.  If  from  the  points  A  and  J5  of  a  straight  line,  and  their  mid-point  C, 
perpendiculars  are  drawn  to  a  given  straight  line,  the  perpendicular  from 
C  is  equal  to  the  half-sum,  or  the  half-difference  of  the  perpendiculars 
from  A  and  i?,  according  as  A  and  B  are  on  the  same  or  on  opposite  sides 
of  the  given  straight  line. 

2.  If  D  and  E  are  the  mid-points  of  the  sides  AB  and  J. (7  of  a  triangle 
ABC,  and  BE  is  produced  to  i^  making  ^ii^  equal  to  BE,  and  CD  is  pro- 
duced to  Q  making  DCr  equal  to  CD,  show  that  G,  A,  and  F  are  in  a 
straight  line. 

Suggestion.     Show  that  DE  is  parallel  to  both  AF  and  AG. 

3.  An  angle  of  a  triangle  is  right,  acute,  or  obtuse  according  as  the 
median  drawn  from  its  vertex  is  equal  to,  greater  than,  or  less  than  half 
the  side  it  bisects. 

4.  The  feet  of  the  four  perpendiculars,  drawn  from  one  vertex  of  a 
triangle  to  the  internal  and  external  bisectors  of  the  other  two  angles,  lie 
on  one  straight  line  which  passes  through  the  mid-points  of  two  sides  of 
the  triangle. 


140-141]       TRIANGLES  AND  PARALLELOGRAMS 


83 


Proposition  XXXIX 

141.  The  straight  lines  drawn  from  the  vertices  of  a 
triangle  perpendicular  to  the  opposite  sides  are  con- 
current. 

A         L 


Let  ABC  be  any  triangle,  and  from  the  three  vertices  draw 
the  straight  lines  AO^y  BO^,  GO^  perpendicular  to  the  opposite 
sides. 


con- 


It  is  required  to  prove  that  AO^  BO2,  and   CO3  are 
current. 

Proof.  Through  A,  B,  and  C  draw  the  straight  lines  KL, 
KM,  and  LM  parallel  respectively  to  the  opposite  sides  of  the 
triangle. 

Then  AC BK  is  a  parallelogram,  as  is  also  ABCL. 

Hence  A  is  the  mid-point  of  KL,  and  AO^  is  perpendicular 
to  KL  since  it  is  perpendicular  to  BC.    (Prop.  XXV.,  Cor.  IV.) 

Similarly  BO2  is  perpendicular  to  KMsit  its  mid-point. 

And  CO3  is  perpendicular  to  LM  at  its  mid-point. 

Therefore  AO^,  BO2,  CO^  are  concurrent.    (Prop.  XXXVIII.) 

Definition.  The  point  X,  in  which  the  three  perpendicu- 
lars from  A,  B,  C,  to  the  opposite  sides  intersect,  is  called  the 


orthocentre  of  the  triangle  ABC. 


84  ELEMENTARY  GEOMETRY  [Chap,  l 


EXERCISES 

1.  If  the  point  D  is  the  orthocentre  of  the  triangle  ABC,  show  that  A  is 
the  orthocentre  of  the  triangle  BCD,  B,  the  orthocentre  of  the  triangle 
ACD,  and  C,  the  orthocentre  of  the  triangle  ABD. 

2.  If  the  opposite  sides  of  a  quadrilateral  are  perpendicular  to  each 
other,  the  diagonals  are  also  perpendicular  to  each  other  (Carnot's 
theorem).     Could  such  a  quadrilateral  be  convex? 


Sectioist  V 

ADDITIONAL  PROBLEMS   OF   CONSTRUCTION 

142.  In  the  preceding  propositions  we  have  introduced  at 
convenient  places  several  problems  of  construction,  not  because 
they  form  an  indispensable  part  of  the  body  of  geometrical 
truth,  but  rather  because  of  their  inherent  interest,  and  because 
they  present  to  the  beginner  matters  for  proof  which  he  clearly 
sees  ought  not  to  be  accepted  without  proof. 

By  making  use  of  the  principles  employed  in  the  preceding 
problems  the  following  can  easily  be  solved : 

1.  To  construct  a  triangle  when  two  sides  and  the  included 
angle  are  given. 

2.  To  construct  a  triangle  when  a  side  and  the  two  adjacent 
angles  are  given. 

3.  To  construct  a  triangle  when  two  angles  and  a  side 
opposite  to  one  of  them  are  given. 

This  problem  can  easily  be  reduced  to  (2)  by  making  use  of 
the  property  of  Proposition  XXVII,  Cor.  I. 

4.  To  construct  a  triangle  when  two  sides  and  an  angle 
opposite  to  one  of  them  are  given. 

The  pupil  should  notice  all  the  cases  that  may  arise  in  this 
problem  according  as  the  given  angle  is  acute,  obtuse,  or  a 
right  angle ;  also  according  as  the  given  angle  is  opposite  the 
greater  or  the  less  of  the  two  given  sides ;  and  should  deter- 
mine under  what  conditions  there  are  two  solutions,  one 
solution,  or  no  solution. 


141-143]        TRIANGLES  AND  PARALLELOGRAMS 


85 


Bv 


^x- 

eA 

/ 

^ 

%<p 

^'\ 

5.  To  trisect  a  right  angle. 

Suggestion.  Upon  any  segment  AG  of  one 
boundary,  terminated  at  one  end  by  the  vertex,  con- 
struct an  equilateral  triangle  AEG.  Similarly  on  an 
equal  segment  AB  construct  an  equilateral  triangle 
ADB.  AD  and  ^^are  the  required  trisectors.  What 
lines  might  be  omitted  in  the  construction  ? 

6.  To  trisect  a  given  segment  of  a  straight  line. 
Suggestion.     Upon  the  given  segment  as  base  construct  an  equilateral 

triangle.     Bisect  the  base  angles.     From  the  point  where  these  bisectors 
meet,  draw  lines  parallel  to  the  sides. 

Typical  Solutions 

143.  We  shall  here  insert  a  few  typical  problems  with  their 
solutions  representing  the  principal  methods  of  attack  when  a 
solution  does  not  readily  appear, 

1.  To  construct  a  triangle  Jiavirig  given  one  side,  an  adjacent 
angle,  and  the  sum  of  the  other  two  sides. 

Let  a  be  equal  to  the  given  side,  b  the  sum  of  the  other  two  sides,  and 
P  the  given  angle. 

Construction.      Draw  any  line  AB  equal  to  a.    At  one  extremity  A 
construct  an  angle  BAG  equal  to  ZP,  and  make  ^C  equal  to  b.    Now 
arises  the    problem  to  find  a  point  D  in  AG  such  that  if  it  be  joined 
to  B,  DB  will  equal  DG,  and  so  the 
sum  of  AD  and  DB  will  equal  AC.  a • 

Assume  the  thing  done.  That  is, 
choose  some  point  D  and  assume  that 
it  fulfils  the  given  conditions. 

Join  DB  and  BG.  Then  BDG  is 
an  isosceles  triangle,  and  a  perpen- 
dicular from  D  upon  BG  would  bi- 
sect BG. 

But  we  were  able  to  draw  BG  with- 
out knowing  D,  also  to  bisect  BG  and 
erect  the  perpendicular  ED,  and  thus 
determine  D. 

So  the  construction  would  be  as  fol- 
lows :  — 


86  ELEMENTARY  GEOMETRY  [Chap.  I 

Draw  AB  and  AC  as  before. 

Join  BC.  Bisect  BC  aX  E  and  draw  the  perpendicular  bisector  cutting 
AC  Sit  D.     Join  DB. 

A  ADB  is  the  required  triangle. 

This  method  of  solution  is  called  Solving  by  Analysis. 

Notice  the  order.  We  first  assumed  the  required  construction  to  have 
been  made,  then  analyzed  the  diagram  and  sought  to  find  upon  what 
principles  the  construction  rests  and  what  steps  would  lead  to  it.  Then 
by  taking  these  steps  in  order  the  solution  is  accomplished. 

2.  Find  a  j)oint  in  a  given  straight  liyie,  which  is  equidistant 
from  tico  given  points,  (1)  on  the  same  side  of  the  line,  (2)  on 
opposite  sides  of  the  line. 

If  A  and  B  are  the  given  points  and  p  the  given  straight  line,  the 
required  point  clearly  is  the  intersection  with  p,  of  the  perpendicular 
bisector  of  AB.  For,  this  perpendicular  bisector  is  the  locus  of  points 
which  are  equidistant  from  A  and  B. 

Thus,  by  considering  the  locus  of  points  satisfying  given  conditions  we 
are  led  to  the  solution  of  the  stated  problem.  This  method  may  be 
called  the  method  of  Intersection  of  Loci. 

3.  Two  points  A  and  B  are  on  opposite  sides  of  a  river  whose 
hanks  are  parallel  straight  lines.  If  the  river  must  be  crossed  in 
a  given  direction,  find  the  shortest  path  that  can  be  taken  from 
A  to  B. 


Let  the  parallel  lines  p  and  q  represent  the  banks  of  the  river,  and  MN 
the  direction  in  which  the  river  must  be  crossed. 

The  line-segment  MN  will  indicate  the  distance  that  roust  be  travelled 
in  the  given  direction. 


143]  TRIANGLES  AND  PARALLELOGRAMS  87 

From  A  draw  a  line  AC  equal  and  parallel  to  MN.  Then  AC  is  the 
distance  that  must  be  travelled  in  the  given  direction ;  the  remaining 
distance  is  to  be  as  short  as  possible,  and  so  will  be  equal  to  the  line- 
segment  CB. 

Let  CB  cut  q  at  E.  Draw  ED  parallel  to  NM  or  CA,  meeting  p  at  D. 
Join  AD.    Then  ADEB  is  the  required  path. 

The  pupil  can  easily  prove  that  A  CED  is  a  parallelogram,  and  hence 
that  the  sum  of  the  distances  ^i>,  DE^  and  EB  is  equal  to  the  sum  of 
the  distances  AC  and  CB.  The  distance  travelled  in  going  from  A  to  B 
would  have  been  the  same  if  we  had  crossed  from  the  line  AD  to  the 
parallel  line  CB.,  in  the  given  direction,  at  any  point,  and  the  only  prob- 
lem was  to  shift  the  given  distance  AC  parallel  to  itself  till  it  reached 
from  one  bank  of  the  river  to  the  other. 

When  a  straight  line  is  moved  so  as  to  remain  always  parallel  to  its 
original  position  it  is  said  to  be  translated,  and  the  motion  is  called  a 
translation. 

When  any  figure  is  moved  so  that  all  of  its  points  describe  parallel 
straight  lines  it  is  said  to  be  translated. 

Since  the  solution  of  this  problem  required  the  translation  of  the  line 
AC,  which  satisfied  some  of  the  conditions  of  the  problem  into  such  a 
position  as  to  satisfy  the  remaining  conditions,  the  method  is  sometimes 
called  the  method  of  Translation. 

EXERCISES 

1.  Through  a  given  point  to  draw  a  straight  line  to  make  equal  angles 
with  two  given  intersecting  lines. 

Suggestion.  Translate  the  bisector  of  the  angle  between  the  lines  till 
it  passes  through  the  given  point.     See  another  solution  on  p.  QQ. 

2.  Construct  a  triangle  having  given  one  side  and  the  orthocentre. 

3.  Construct  a  triangle  having  given  the  base,  the  altitude,  and  the 
length  of  one  side.    Is  the  solution  always  possible  ? 

4.  Through  a  given  point  P  draw  a  straight  line  such  that  the  perpen- 
diculars to  it  from  two  fixed  points  meet  it  at  equal  distances  from  P. 

5.  Through  a  given  point  draw  a  line-segment  terminated  by  two  given 
intersecting  straight  lines,  which  shall  be  bisected  at  the  given  point. 

6.  Construct  a  right  triangle  having  given  the  hypotenuse  and  the 
difference  of  the  two  sides. 

7.  Construct  a  triangle  equiangular  with  a  given  triangle  and  having  a 
given  perimeter. 


88 


ELEMENTARY  GEOMETRY 


[Chap.  I 


Section  VI 


SYMMETRICAL   FIGURES 


144.   If  there  be  given  any  straight  line  a,  and  from  a  point 
A  we  draw  AM  perpendicular  to  the  given  line,  and  produce  it 
to  a  point  A'  such  that  MA  equals  MA,  then  A  and  A'  are 
called  inverse  points  relative  to  the 
line  a. 

In  the  figure,  B  and  B^  form  an- 
other pair  of  inverse  points  relative 
to  a. 

Definition.  If  two  figures  are 
such  that  for  every  point  of  the  one 
there  is  an  inverse  point  in  the  other, 
relative  to  a  certain  straight  line, 
the  two  figures  are  said  to  be  sym- 
metrical with  respect  to  this  line, 
and  the  line  is  called  an  axis  of  sym- 
metry for  the  two  figures. 


a 
4. 4' 

M 

B 
A 

A 

A 

A 

C            B 

B'             C 

145.  If  you  think  of  the  plane  in  which  there  lie  two  figures 
symmetrical  with  respect  to  a  certain  straight  line,  as  being 
folded  along  this  line,  just  as  you  would  fold  a  sheet  of  paper 
along  a  given  straight  line,  it  is  clear  that  the  two  figures  would 
come  to  coincide,  since  all  lines  perpendicular  to  the  axis  of 
symmetry  would  fold  over  upon  themselves. 

Hence, 

Theorem.  If  two  figures  are  symmetrical  with  respect  to  a 
straight  line,  they  are  superposable  by  inversion. 


A  figure  symmetrical  with  a  given  figure  can  easily  be  obtained 
by  tracing  it  with  ink  and  then  folding  the  paper  along  a 
straight  line  before  the  ink  is  dry. 


144-146]       TRIANGLES  AND  PARALLELOGRAMS 


89 


146.  Two  parts  of  the  same  figure  may  be 
symmetrical  with  respect  to  a  given  straight 
line,  as  in  the  diagram.  In  such  cases  we  say 
that  the  figure  is  symmetrical  with  respect  to 
the  line. 

Theorem.  A  circle  is  symmetrical  with  respect 
to  any  of  its  diameters. 

Let  APB  be  a  circle  whose  centre  is  0,  AB 
any  diameter,  P  any  point  on  the  circle. 

From  P  draw  PM  perpendicular  to  the  diame- 
ter, and  produce  it  to  meet  the  circle  again  at  P'.     Join  PO 
and  P'O. 


The  As  0PM  and  0PM  are  identically  equal.  Hence  P  is 
the  inverse  point  of  P,  and  for  every  point  of  the  circle  there 
is  an  inverse  point  with  respect  to  the  diameter. 


EXERCISES 

1.  Show  that  a  square  is  symmetrical  with  respect  to  each  of  its 
diagonals. 

2.  Show  that  an  equiangular  triangle  is  symmetrical  with  respect  to 
each  of  its  medians. 

3»    If  a  quadrilateral  is  symmetrical  with  respect  to  one  of  its  diagonals, 
show  that  the  two  diagonals  are  at  right  angles. 

4.   If  a  triangle  is  symmetrical  with  respect  to  a  median,  it  must  be 
isosceles. 


90  ELEMENTARY  GEOMETRY  [Chap.  I 


MISCELLANEOUS  EXERCISES 

1.  In  a  quadrilateral  ABCD,  if  AB  is  the  greatest  side  and  CD  the 
least,  show  that  A  BCD  is  greater  than  Z  DAB,  and  Z  CD  A  greater  than 
ZABC. 

2.  Show  that  the  sum  of  the  diagonals  of  a  convex  quadrilateral  is 
greater  than  the  sum  of  either  pair  of  opposite  sides,  and  also  greater 
than  half  the  sum  of  the  four  sides. 

3.  The  sum  of  the  distances  of  any  point  within  a  rectilinear  figure 
from  the  vertices  is  greater  than  half  the  sum  of  the  sides. 

4.  If  D  is  any  point  on  the  side  BC  of  &  triangle  ABC,  show  that  the 
sum  of  the  sides  of  the  triangle  is  greater  than  twice  AD. 

5.  Show  that  no  convex  polygon  can  have  more  than  three  of  its 
interior  angles  acute,  nor  more  than  three  of  its  exterior  angles  obtuse. 

6.  On  a  given  straight  line  find  a  point  which  is  equidistant  from  two 
given  lines.     Under  what  conditions  is  the  solution  impossible  ? 

7.  Show  how  to  find  a  point  which  is  at  a  given  distance  from  each 
of  two  intersecting  straight  lines.     How  many  such  points  are  there  ? 

8.  If  two  quadrilaterals  are  mutually  equiangular,  and  two  adjacent 
sides  of  the  one  are  respectively  equal  to  the  two  corresponding  adjacent 
sides  of  the  other,  show  that  the  quadrilaterals  are  identically  equal. 

9.  Show  that  the  sum  of  any  two  medians  of  a  triangle  is  greater 
than  the  third  median. 

10.  Show  that  the  sum  of  the  three  medians  of  a  triangle  is  greater 
than  three-fourths  of  the  sum  of  the  sides  of  the  triangle. 

11.  Show  that  if  two  of  the  medians  of  a  triangle  are  equal,  the  tri- 
angle must  be  isosceles. 

12.  If  E  and  F  are  the  mid-points  of  the  sides  AB  and  CD  of  the 
parallelogram  ABCD,  show  that  the  lines  ED  and  BF  will  trisect  the 
diagonal  AC. 

13.  If  the  vertices  of  one  parallelogram  lie  on  the  sides  of  another, 
show  that  all  four  diagonals  pass  through  the  same  point. 

14.  If  an  exterior  angle  of  a  triangle  be  bisected,  and  also  one  of  the 
interior  non-adjacent  angles,  the  angle  made  by  the  two  bisectors  is  equal 
to  half  of  the  other  interior  non-adjacent  angle. 


146]  TRIANGLES  AND  PARALLELOGRAMS  91 

16.  If  two  triangles  on  the  same  side  of  a  common  base  have  their 
sides  which  are  terminated  in  opposite  extremities  of  the  base  equal,  the 
line  joining  the  vertices  will  be  parallel  to  the  common  base. 

16.  From  the  vertex  A  of  any  triangle  ABC  two  straight  lines  are 
drawn  meeting  the  opposite  side  at  D  and  E,  so  as  to  make  the  angle 
BAD  equal  to  the  angle  O,  and  the  angle  CAE  equal  to  the  angle  B. 
Show  that  the  triangle  DAE  is  isosceles. 

17.  In  any  triangle  ABC,  the  bisector  of  the  angle  A  makes  wit'h  the 
perpendicular  drawn  from  A  to  the  opposite  side  an  angle  equal  to  half 
the  difference  between  the  angles  B  and  C. 

18.  In  any  triangle  ABC,  the  bisectors  of  the  angles  B  and  C  make 
with  each  other  an  angle  greater  by  a  right  angle  than  one-half  A. 

19.  In  any  triangle  ABC,  the  bisectors  of  the  exterior  angles  at  B  and 
C  make  with  each  other  an  angle  less  by  a  right  angle  than  one-half  A. 

20.  In  any  convex  quadrilateral,  the  bisectors  of  two  consecutive  angles 
make  with  each  other  an  angle  equal  to  the  half-sum  of  the  other  two 
angles. 

21.  If  through  the  point  of  intersection  of  the  bisectors  of  the  angles  B 
and  C  in  any  triangle  ABC,  a  line-segment  MN  is  drawn  parallel  to  the 
side  BC,  and  terminated  by  the  sides  AB  and  AC,  show  that  3IN  equals 
the  sum  of  BM  and  CN. 

22.  If  upon  one  boundary  OX  of  an  angle  you  choose  two  points  A 
and  A',  and  upon  the  other  boundary  OF  you  choose  two  points  B  and 
B',  such  that  OB  equals  OA  and  OB'  equals  OA',  and  join  crosswise  AB' 
and  A'B,  show  that  the  intersection  of  these  lines  is  upon  the  bisector 
of  the  angle  XO  Y. 

23.  If  two  sides  of  a  triangle  are  unequal,  the  median  through  their 
intersection  makes  the  greater  angle  with  the  lesser  side. 

24.  A  quadrilateral  which  has  two  sides  equal  and  the  other  two  sides 
parallel  may  be  a  parallelogram,  or  it  may  be  a  trapezoid  having  an  axis 
of  symmetry. 

25.  In  an  isosceles  triangle  the  sum  of  the  distances  of  a  point  in  the 
base  from  the  two  sides  is  the  same,  no  matter  where  in  the  base  the  point 
is  chosen. 

If  the  point  should  be  chosen  in  the  base  produced,  how  would  this 
theorem  be  altered  ? 

26.  In  an  equilateral  triangle  the  sum  of  the  distances  of  a  point  within 
the  triangle  from  the  three  sides  is  the  same,  no  matter  where  the  point 
is  chosen. 


92  ELEMENTARY  GEOMETRY  [Chap.  I 

SUMMARY  OF   CHAPTER  I 
1.   Definitions. 

(1)  Scalene   Triangle,    Isosceles   Triangle,    Equilateral   Triangle. 

See  §  33. 

(2)  Right  Triangle  —  a  triangle  having  one  right  angle.     §  61. 

(3)  Hypotenuse  of  a  Right   Triangle  —  the  side  opposite  the  right 

angle.     §  61. 

(4)  Complementary  Angles  —  two  angles  whose  sum  is  equal  to  a 

right  angle.     §  102. 

(5)  Corollary  —  a  theorem,  the  truth  of  which  is  easily  deduced  from 

another,  or  the  proof  of  which  is  easily  deduced  from  the  proof 
of  another.     §  50. 

(6)  Converse  Theorems  —  two  theorems  such  that  the  hypothesis  of 

each  is  the  conclusion  of  the  other.     §  44. 

(7)  Locus  —  a  line  or  group  of  lines  such  that  all  their  points  satisfy 

certain  conditions  while  no  other  points  satisfy  those  conditions. 
§73. 

(8)  Parallel  Lines  —  straight  lines  lying  in  the  same  plane,  and  such 

that  if  they  are  drawn  ever  so  far  either  way  they  will  not  in- 
tersect.    §  85. 

(9)  Polygon  —  a  closed  rectilinear  figure  of  any  number  of  sides. 

§114. 

(10)  Convex  Polygon  —  one  which  lies  wholly  on  one  side  or  the  other 

of  each  of  its  sides.     §  115. 

(11)  Diagonal  of  a  Polygon  —  a  straight  line  joining  any  two  non- 

consecutive  vertices.     §  109. 

(12)  Quadrilateral  —  a  closed   rectilinear  figure  having   four  sides. 

§114. 

(13)  Parallelogram  —  a  quadrilateral  whose  opposite  sides  are  parallel. 

§119. 

(14)  Trapezoid  —  a  quadrilateral  having  one  and  only  one  pair  of  op- 

posite sides  parallel.     §  119. 

(15)  Isosceles  Trapezoid  —  a  trapezoid  having  its  non-parallel  sides 

equal.     Ex.  7,  p.  77. 

(16)  Trapezium  —  a  quadrilateral  having  no  two  sides  parallel.    §  119. 

(17)  Rhombus  —  a  parallelogram   having  two  adjacent  sides  equal. 

§120. 

(18)  Rectangle  —  a  parallelogram  having  one  right  angle.     §  121. 

(19)  Square  —  a  parallelogram  having  one  right  angle  and  a  pair  of 

adjacent  sides  equal.     §  122. 


Summary]     TRIANGLES  AND  PARALLELOGRAMS  93 

(20)  Perimeter  of  a  Polygon  —  the  sum  of  its  sides.     §  132. 

(21)  Median  of  a  Triangle  —  a  straight  line  drawn  from  any  vertex 

to  the  mid-point  of  the  opposite  side.     §  134. 

(22)  Centroid  of  a  Triangle  —  the  point  at  which  its  medians  intersect. 

§137. 

(23)  Concurrent  Lines  —  straight  lines  which  pass  through  the  same 

point.     §  135. 

(24)  Inverse  Points  —  two  points  are  inverse  relative  to  a  given  straight 

line  when  the  line-segment  joining  them  is  bisected  perpendicu- 
larly by  the  given  line.     §  144. 

(25)  Symmetrical  Figures  —  a  figure  is  symmetrical  relative  to  a  given 

straight  line  when  for  every  point  of  the  figure  there  is  an  in- 
verse point  of  the  figure  relative  to  that  line.     §  144. 

Two  figures  are  symmetrical  relative  to  a  given  straight  line 
when  for  every  point  of  one  figure  there  is  an  inverse  point  of 
the  other,  relative  to  that  line.     §  144. 

(26)  Axis  of  Symmetry  —  a  straight  line  relative  to  which  a  figure  is 

symmetrical.     §  144. 

2.  Axioms. 

Two  magnitudes  A  and  B  of  the  same  kind  are  either  equal,  or  A  is 
greater  than  B,  or  A  is  less  than  B  (Axiom  10).     §  64. 

3.  Postulates. 

Through  any  point  one  and  only  one  straight  line  can  be  drawn  parallel 
to  a  given  straight  line  (Postulate  5).     §  86. 

4.  Problems. 

(1)  To  construct  an  equilateral  triangle,  having  given  the  length  of 

each  side.     §  34. 

(2)  To  construct  an  isosceles  triangle,  having  given  the  lengths  of 

the  sides.     §  37. 

(3)  To  construct  any  triangle,  having  given  the  lengths  of  the  sides. 

§39. 

(4)  To  bisect  a  given  angle.     §  54. 

(5)  To  bisect  a  given  line-segment.     §  56. 

(6)  Prom  a  given  point  in  a  straight  line  to  draw  a  perpendicular  to 

the  line.     §71. 

(7)  From  a  given  point  without  a  given  straight  line,  to  draw  a  per- 

pendicular to  the  line.     §  75. 

(8)  At  a  given  point  on  a  given  straight  line  to  construct  an  angle 

equal  to  a  given  angle.     §  84. 


94  ELEMENTARY  GEOMETRY  [Chap.  I 

(9)  Through  a  given  point  to  draw  the  straight  line  parallel  to  a 
given  straight  line.     §  99. 

(10)  To  construct  a  triangle  having  given  one  side,  an  adjacent  angle, 

and  the  sum  of  the  other  two  sides.     §  143.  1. 

(11)  In  a  given  straight  line  to  find  a  point  equidistant  from  two 

given  points.     §  143.  2. 

(12)  To  find  the  shortest  path  between  two  given  points  when  a  part 

of  the  path  is  given  both  in  length  and  direction.     §  143.  3. 

5.  Theorems  on  Bisectors. 

(1)  Through  the  vertex  of  an  angle  there  can  be  drawn  one  and  only 

one  straight  line  which  bisects  the  angle.    §  46. 

(2)  In  a  given  line-segment  there  can  be  found  one  and  only  one 

point  which  bisects  the  line-segment.     §  47. 

(3)  Two  points  each  equidistant  from  the  extremities  of  a  given 

line-segment  determine  the  perpendicular  bisector  of  the  line- 
segment.     §  57. 

(4)  The  bisector  of  an  angle  is  the  locus  of  points  within  the  angle 

which  are  equidistant  from  its  boundaries.     §  107. 

6.  Theorems  on  Perpendiculars. 

(1)  At  any  point  in  a  straight  line  there  can  be  drawn  one,  and  only 

one,  perpendicular  to  the  line.     §  46. 

(2)  From  a  point  outside  a  given  straight  line  not  more  than  one 

perpendicular  to  the  line  can  be  drawn.     §  62. 

(3)  Perpendiculars  to  the  same  straight  line,  lying  in  the  same  plane, 

are  parallel.     §  91. 

(4)  If  a  straight  line  is  perpendicular  to  one  of  two  parallel  lines,  it 

is  perpendicular  also  to  the  other.     §  97. 
(6)  If  two  straight  lines  intersect,  lines  perpendicular  to  them  in  the 
same  plane  will  also  intersect.     §  98. 

7.  Theorems  on  the  Equality  of  Triangles. 

Two  triangles  are  identically  equal  if  they  have  — 

(1)  Two  sides  and  the  included  angle  of  one  equal,  respectively,  to 

two  sides  and  the  included  angle  of  the  other.     §  41. 

(2)  A  side  and  the  two  adjacent  angles  of  one  equal,  respectively,  to 

a  side  and  the  two  adjacent  angles  of  the  other.     §  43. 

(3)  The  three  sides  of  one  equal,  respectively,  to  the  three  sides  of 

the  other.     §  53. 

(4)  Each  a  right  angle,  and  the  hypotenuse  and  a  side  of  one  equal, 

respectively,  to  the  hypotenuse  and  a  side  of  the  other.     §  78. 


Summary]     TBIANGLES  AND  PARALLELOGRAMS  95 

(5)  Two  sides  of  one  equal,  respectively,  to  two  sides  of  the  other 

and  the  angles  opposite  a  pair  of  equal  sides  also  equal 
§§81,82. 

(6)  Two  angles  of  one  equal,  respectively,  to  two  angles  of  the  other, 

and  the  sides  opposite  one  pair  of  equal  angles  also  equal. 
§106. 

8.  Theorems  on  Unequal  Triangles. 

(1)  If  two  triangles  have  two  sides  of  the  one  equal,  respectively,  to 

two  sides  of  the  other,  and  the  included  angles  unequal,  the 
triangle  which  has  the  greater  included  angle  has  also  the 
greater  third  side.     §  79. 

(2)  If  two  triangles  have  two  sides  of  the  one  equal,  respectively,  to 

two  sides  of  the  other,  and  the  third  sides  unequal,  the  triangle 
which  has  the  greater  third  side  has  also  the  greater  included 
angle.     §  80. 

9.  Theorems  on  the  Properties  of  Triangles. 

(1)  In  an  isosceles  triangle,  the  angles  opposite  the  equal  sides  are 

equal.     §  48. 

(2)  The  straight  line  which  bisects  the  vertical  angle  of  an  isosceles 

triangle  also  bisects  the  base,  and  is  perpendicular  to  the 
base.     §  49. 

(3)  An  equilateral  triangle  is  also  equiangular.     §  50. 

(4)  If  two  angles  of  a  triangle  are  equal,  the  sides  opposite  those 

angles  are  also  equal.     §  51. 

(5)  An  equiangular  triangle  is  also  equilateral.     §  52, 

(6)  If  one  side  of  a  triangle  is  produced,  the  exterior  angle  so  formed 

is  greater  than  either  of  the  two  interior  non-adjacent  angles. 
§59. 

(7)  Any  two  angles  of  a  triangle  are  together  less  than  two  right 

angles.     §  00. 

(8)  If  a  triangle  has  one  right  angle  or  one  obtuse  angle,  the  othei 

two  angles  must  be  acute.     §  61. 

(9)  If  two  sides  of  a  triangle  are  unequal,  the  angle  opposite  the  greater 

side  is  greater  than  the  angle  opposite  the  less.     §  63. 

(10)  If  two  angles   of   a    triangle    are   unequal,    the   side    opposite 

the  greater  angle  is  greater  than  the  side  opposite  the  less. 
§65. 

(11)  In  a  right  triangle  the  hypotenuse  is  the  greatest  side.     §  66. 


96  ELEMENTARY  GEOMETRY  [Chap.  I 

(12)  Any  two  sides  of  a  triangle  are  together  greater  than  the  third 

side.     §  70. 

(13)  If  one  side  of  a  triangle  is  produced,  the  exterior  angle  so  formed 

is  equal  to  the  sum  of  the  two  interior  non-adjacent  angles. 
§100. 

(14)  The  sum  of  the  three  angles  of  a  triangle  is  equal  to  two  right 

angles.    §  101. 

(15)  The  line-segment  joining  the  mid-points  of  two  sides  of  a  triangle 

is  parallel  to  the  third  side  and  equal  to  half  of  it.     §  130. 

(16)  If  from  the  mid-point  of  one  side  of  a  triangle  there  is  drawn  a 

straight  line  parallel  to  a  second  side,  this  line  passes  through 
the  mid-point  of  the  third  side.     §  133. 

(17)  The  medians  of  a  triangle  are  concurrent.     §  136. 

(18)  The  centroid  of  a  triangle  is  a  trisection  point  of  each  median. 

§138. 

(19)  The  bisectors  of  the  angles  of  a  triangle  are  concurrent.     §  139. 

(20)  The  perpendicular  bisectors  of  the  sides  of  a  triangle  are  con- 

current.    §  140. 

(21)  The  straight  lines  drawn  from  the  vertices  of  a  triangle  perpen- 

dicular to  the  opposite  sides  are  concurrent.     §  141. 

10.  Theorems  on  Parallel  Lines. 

(1)  If  two  straight  lines  lying  in  one  plane  are  such  that  a  transversal 

makes  a  pair  of  alternate  angles  equal,  or  any  pair  of  corre- 
sponding angles  equal,  these  two  lines  are  parallel.     §§  88,  89. 

(2)  If  two  straight  lines  lying  in  one  plane  are  such  that  a  transversal 

makes  a  pair  of  interior  angles,  or  a  pair  of  exterior  angles,  on 
the  same  side  of  it  supplementary,  these  two  lines  are  parallel. 
§90. 

(3)  Two  straight  lines  parallel  to  the  same  straight  line  are  parallel  to 

each  other.     §  92. 

(4)  If  two  parallel  straight  lines  are  cut  by  a  transversal,  any  two 

alternate  angles,  and  any  two  correspojiding  angles,  are  equal. 
§§  93,  94. 

(5)  If  two  parallel  straight  lines  are  cut  by  any  transversal,  the  interior 

angles,  and  also  the  exterior  angles,  on  the  same  side,  are 
supplementary.     §  96. 

11.  Theorems  on  Convex  Polygons. 

(1)  The  sum  of  the  interior  angles  of  any  convex  polygon,  together 
with  four  right  angles,  makes  up  twice  as  many  right  angles  as 
the  polygon  has  sides.     §  116. 


Summary]     TRIANGLES  AND  PARALLELOGRAMS  97 

(2)  If  the  sides  of  a  convex  polygon  are  produced  in  order,  the 
exterior  angles  so  formed  are  together  equal  to  four  right 
angles.     §  117. 

12.  Theorems  on  Parallelograms. 

(1)  The  opposite  sides  and  the  opposite  angles  of  a  parallelogram  are 

equal.     §  123. 

(2)  Either  diagonal  of  a  parallelogram  divides  the  figure  into  two 

superposable  triangles.     §  124. 

(3)  The  diagonals  of  a  parallelogram  bisect  each  other.     §  126. 

(4)  If  two  sides  of  a  quadrilateral  are  parallel  and  equal,  the  figure 

is  a  parallelogram.     §  127. 

(5)  If  both  pairs  of  opposite  sides  of  a  quadrilateral  are  equal,  the 

figure  is  a  parallelogram.     §  128. 

(6)  If  two  parallelograms  have  two  adjacent  sides  and  the  included 

angle  of  one  equal,  respectively,  to  two  adjacent  sides  and  the 
included  angle  of  the  other,  the  parallelograms  are  identically 
equal.     §  129. 

13.  Miscellaneous  Theorems. 

(1)  If  two  straight  lines  intersect,  the  vertical  angles  are  equal.    §  58. 

(2)  Not  more  than  two  equal  line-segments  can  be  drawn  from  a  given 

point  to  a  given  straight  line.     §  67. 

(3)  If  any  pair  of  alternate  angles  formed  by  two  straight  lines  and 

a  transversal  are  unequal,  or  if  any  pair  of  corresponding  angles 
are  unequal,  or  if  the  interior  angles  on  one  side  are  not  sup- 
plementary, the  two  lines  are  not  parallel,  and  therefore  will 
meet  if  produced.     §  96. 

(4)  If  the  boundaries  of  one  angle  are  respectively  parallel  or  per- 

pendicular to  the  boundaries  of  another,  these  two  angles  are 
either  equal  or  supplementary.     §  105. 

(5)  If  the  mid-points  of  the  sides  of  any  quadrilateral  be  joined  in 

order,  the  figure  so  formed  is  a  parallelogram,  and  the  sum  of 
the  sides  of  this  parallelogram  equals  the  sum  of  the  diagonals 
of  the  quadrilateral.     §  131. 

14.  On  Symmetry. 

(1)  If  two  figures  are  symmetrical  with  respect  to  a  straight  line, 

they  are  superposable  by  inversion.     §  145. 

(2)  A  circle  is  symmetrical  with  respect  to  any  of  its  diameters. 

§146. 


CHAPTER   II 
THE   CIRCLE 

Section  I 

DEFINITIONS   AND   PRELIMINARY  THEOREMS 

147.  In  the  introductory  chapter  the  following  definitions 
were  given : 

A  circle  is  a  closed  line  all  points  of  which  are  equally 
distant  from  a  certain  point  within  it  called  the  centre. 

The  line-segment  joining  the  centre  to  any  point  of  the  circle 
is  called  a  radius,  and  a  line-segment  through  the  centre 
terminated  both  ways  by  the  circle  is  called  a  diameter. 

From  these  definitions  it  follows  that  all  radii  of  the  same 
circle  are  equal,  and  all  diameters  of  the  same  circle  are  equal. 

The  distance  from  the  centre  of  any  point  inside  the  circle 
is  less  than  a  radius,  and  of  any  point  outside  the  circle  is 
greater  than  a  radius. 

148.  Theorem  I.  Two  circles  in  a  plane  which  have  the 
same  centre  and  equal  radii  coincide  throughout. 

For,  if  there  is  any  point  of  one  circle  which  does  not 
coincide  with  a  point  of  the  other,  it  must  lie  either  inside  or 
outside  of  the  other,  and  hence  its  distance  from  the  centre  is 
either  less  or  greater  than  a  radius  of  the  other.  But  this  is 
not  the  case,  since  all  radii  of  the  same  circle  are  equal  by 
definition  and  radii  of  the  two  circles  are  assumed  to  be  equal. 

Corollary  I.  Two  circles  in  a  plane  which  have  the  same 
centre  and  one  point  in  common  coincide  throughout. 

98 


Chap.  II,  147-151]  THE  CIRCLE  99 

For  they  have  the  same  centre  and  equal  radii,  since  the 
line  joining  their  common  point  to  the  centre  is  a  radius  of 
each  circle. 

Corollary  II.  Two  circles  which  have  equal  radii  can  he 
made  to  coincide,  and  hence  are  identically  equal;  and  conversely, 
equal  circles  have  equal  radii. 

149.  Theorem  II.  Through  a  given  point  any  number  of 
different  circles  can  he  described. 

For,  if  A  is  the  given  point,  we  may  choose  any  point  0 
whatever  for  centre,  and  with  radius  OA  describe  a  circle 
which  passes  through  A. 

Since  a  circle  is  a  closed  curve,  two  circles  which  intersect 
at  one  point  must  also  intersect  at  a  second  point.       (Art.  25.) 

150.  Theorem  III.  Through  two  given  points  any  number 
of  different  circles  can  be  described. 

Let  A  and  B  be  the  given  points;  then  every  point  of 
the  perpendicular  bisector  of  the  line-segment  AB  is  equi- 
distant from  A  and  B  (Art.  72).  Choose  any  point  0  on  this 
perpendicular  bisector  for  centre,  and  the  circle  described  with 
radius  OA  will  pass  through  both  A  and  B. 

151.  Theorem  IV.  Through  three  given  points  not  in  the 
same  straight  line,  one  and  only  one  circle  can  be  described. 

Let  A,  B,  and  C  be  the  three  given  points.  Then  the  locus 
of  points  equidistant  from  A  and  B  is  the  perpendicular  bisector 
of  the  line-segment  AB,  and  the  locus  of  points  equidistant 
from  B  and  C  is  the  perpendicular  bisector  of  the  line-seg- 
ment BC.  These  two  loci  have  one  (Art.  98)  and  only  one 
common  point ;  which  point,  0  say,  is  equidistant  from  A,  B, 
and  C,  and  is  the  only  such  point. 

With  0  as  centre  and  with  radius  OA  a  circle  may  be 
described  through  A,  B,  and  C,  while  no  other  circle  can  be 


100  ELEMENTARY  GEOMETRY  [Chap.  II 

described  through  these  three  points,  since  no  other  point  than 
0  can  be  found  for  centre. 

Corollary  I.  Two  circles  which  coincide  at  three  points 
coincide  throughout. 

For  they  must  have  the  same  centre  and  equal  radii. 

Corollary  II.  Two  different  circles  can  have  at  most  tivo 
points  in  common; 

Or,  two  circles  can  intersect  in  at  most  tivo  points. 

Incidentally  in  this  article  we  have  solved  the  problem : 

To  find  the  centre  of  the  circle  which  passes  through  three 
given  points,  or  of  which  at  least  three  points  are  given. 

This  is  the  same  problem  as  To  pass  a  circle  through  the  three 
vertices  of  a  triangle,  a  thing  which  can  always  be  done. 

152.  The  questions  arise: 

Can  a  circle  be  described  through  three  given  points  which  lie 
in  a  straight  line  ? 

If  you  should  proceed  as  in  the  case  where  the  three  points  do 
not  lie  in  one  straight  line,  in  what  particular  would  the  con- 
struction fail  ? 

How  many  points  of  a  straight  line  are  equidistant  from  any 
given  point  ?  (See  Art.  67.) 

Theorem  V.  A  straight  line  can  intersect  a  circle  in  at  most 
two  points; 

Or,  what  is  the  same  thing,  a  straight  line  can  have  at  most 
tico  points  in  common  with  any  circle, 

153.  Definitions.  Any  portion  of  a  circle  terminated  by 
two  points  is  called  an  arc  of  the  circle. 

The  straight  line  joining  any  two  points  of  a  circle,  i.e. 
joining  the  extremities  of  an  arc,  is  called  a  chord  of  the  circle. 


151-156]  TBE  CIRCLE  101 

The  chord  is  said  to  subtend  the  arc,  the  a7C  to  bo  .subtended 
by  the  chord. 

Every  chord  subtends  two  arcs  of  the  circle,  one  on  either 
side  of  it,  called  the  greater  or  major  arc  and  the  lesser  or  minor 
arc. 

The  two  arcs  subtended  by  any  chord  together  make  up 
the  whole  circle.  Each  of  these  arcs  is  called  the  conjugate 
of  the  other. 

When  we  speak  of  the  arc  subtended  by  a  given  chord  we 
shall  always  have  in  mind  the  lesser  or  minor  arc  unless  the 
contrary  is  expressly  stated. 

154.  The  arc  of  a  circle  subtended  by  a  diameter  is  called  a 
semicircle. 

Since  a  circle  is  symmetrical  with  respect  to  any  of  its 
diameters  (Art.  146),  a  semicircle  and  its  conjugate  semicircle 
are  equal.  If  you  should  fold  the  circle  over,  along  the 
diameter,  the  two  semicircles  would  coincide. 

Hence  any  semicircle  is  half  of  a  circle,  and  all  semicircles 
belonging  to  the  same  or  equal  circles  are  equal. 

155.  Theorem  VI.  The  perpendicular  bisector  of  any  chord 
of  a  circle  passes  through  the  centre. 

For  it  is  the  locus  of  points  equidistant  from  the  extremities 
of  the  chord,  and  the  centre   being  equidistant  from  the 
•  extremities  of  the  chord  must  lie  on  this  locus. 

156.  Definitions.  The  figure  formed  by  an  arc  of  a 
circle  and  its  subtending  chord  is  called  a  segment  of  the 
circle. 

The  figure  formed  by  an  arc  and  the  two 
ladii  to  its  extremities  is  called  a  sector  of 
the  circle.  The  angle  at  the  centre  formed 
by  two  radii  is  called  the  angle  at  the  centre 
subtended  by  the  intercepted  arc,  or  by  the 
chord  of  that  arc. 


l02  ■BLEUENTABY  GEOMETRY  [Chap.  II 

Proposition  I 

157.  If  in  equal  circles  two  angles  at  the  centre  are 
equal,  the  arcs  subtending  them  are  also  equal;  and  of 
two  unequal  angles  at  the  centre,  the  greater  is  sub- 
tended by  the  greater  arc. 


In  the  equal  circles  AiB^H^  and  A2B2H2,  of  which  the  centres 
are  Oj  and  O2, 

First,  let  the  angle  A^i^i  be  equal  to  the  angle  ^2^2-^2- 

It  is  required  to  prove  that  the  arc  A^Bi  is  equal  to  the  arc 
A2B2. 

Proof.  Superpose  the  circle  Oi  on  the  circle  O2  so  that  their 
centres  coincide.  Then  the  circles  will  coincide  throughout, 
since  by  hypothesis  they  are  equal,  and  every  point  of  the  one 
will  coincide  with  a  point  of  the  other. 

Further,  make  Z  A^O^Bi  coincide  with  its  equal  Z  A2O2B2. 

Since  the  radii  of  the  two  circles  are  equal,  Ai  will  coincide 
with  A2,  and  B^  with  B2,  so  that  the  arc  AiBi  will  coincide 
with  and  be  equal  to  the  arc  ^2-^2. 

Next,  let  the  angle  A^O^B^  be  greater  than  the  angle  ^2^2-52- 
It  is  required  to  prove  that  the  arc  A^B^  is  greater  than  the 
arc  ^2^2- 

Proof.  As  before,  superpose  the  circle  Oj  on  the  circle  O2  so 
that  their  centres  coincide,  and  so  that  the  radius  OiA^  coin- 
cides with  the  radius  02^2- 


157-160]  THE  CIRCLE  103 

Since  /.  AiOiBi  is  greater  than  /.  AiOA^  the  radius  OiB^ 
will  not  coincide  with  O2B2,  but  will  fall  outside  Z  A2O2B2  in 
some  position  such  as  O2B. 

The  arc  A^B^,  which  will  coincide  with  the  arc  A^B,  will 
therefore  overlap  and  be  greater  than  the  arc  A2B2. 

158.  Corollary.  If  in  the  same  circle  two  angles  at  the 
centre  are  equal,  the  arcs  subtending  them  are  also  equal;  and  of 
two  unequal  angles  at  the  centre,  the  greater  is  subtended  by  the 
greater  arc. 

Proposition  II 

159.  //  in  equal  circles  two  arcs  are  equal,  the  angles 
at  the  centre  which  they  subtend  are  also  equal;  and  of 
two  unequal  arcs  the  greater  subtends  the  greater  angle 
at  the  centre. 

[This  theorem  is  the  converse  of  the  theorem  in  Proposition  I, 
and  may  be  proved  either  indirectly  or  by  superposition. 
The  pupil  should  write  out  the  proof  by  both  methods.] 

160.  Corollary.  If  in  the  same  circle  two  arcs  are  equal, 
the  angles  at  the  centre  which  they  subtend  are  also  equal ;  and  of 
two  unequal  arcs,  the  greater  subtends  the  greater  angle  at  the 
centre. 

EXERCISES 

1.  Every  point  of  a  chord  of  a  circle,  except  its  extremities,  lies  inside 
the  circle. 

Let  O  be  the  centre  of  the  circle,  A  and  B  the  extremities  of 
the  chord,  E  any  other  point  of  the  chord.  Prove  that  OE 
is  less  than  OA. 

2.  Find  the  centre  of  a  circle  having  given  any  arc  of  it. 

3.  What  is  the  locus  of  the  centres  of  circles  passing  through  two 
given  points  ? 

4.  Describe  a  circle  that  shall  pass  through  two  given  points  and  have 
a  given  radius. 


104  ELEMENTARY  GEOMETRY  [Chap.  II 


Proposition  III 

161.  //  in  the  same  circle  or  in  equal  circles  two  arcs 
are  equal,  the  chords  subtending  them  are  also  equal; 
and  of  two  unequal  minor  arcs,  the  greater  is  subtended 
by  the  greater  chord. 


Let  the  pupil  give  a  particular  enunciation  of  the  theorem, 
applying  it  to  the  diagram. 

Suggestions  for  Proof.  First,  (a)  equal  arcs  subtend  equal 
angles  at  the  centre.  (Prop.  II.) 

(6)  As  OiA^Bi  and  O^A^B.,  are  identically  equal.       (Art.  41.) 
(c)  Therefore  the  chord  A^B^  equals  the  chord  ^2^2- 
Next,  (a)  the  greater  arc  subtends  the  greater  angle  at  the 
centre.  (Prop.  II.) 

(6)  Apply  Proposition   XX,  Ch.  I,  to   the   As  OiA^B^  and 

C'2-^2-^2* 

Therefore  ... 

162.  Corollary.  If  in  the  same  circle,  or  in  equal  circles, 
two  angles  at  the  centre  are  equal,  the  chords  subtending  them  are 
also  equal;  and  of  two  unequal  angles  at  the  centre,  both  less  than 
straight  angles,  the  greater  is  subtended  by  the  greater  chord. 

Exercise.  —  Why  is  it  necessary  in  this  proposition  and  its  corollary  to 
assume  that  the  unequal  arcs  are  minor  arcs,  and  that  the  unequal  angles 
are  less  than  straight  angles  ? 


161-165] 


THE  CIRCLE 


105 


Proposition  IV 

163.  //  in  the  same  circle,  or  in  equal  circles,  two 
chords  are  equal,  the  arcs  subtended  by  them  are  also 
equal ;  and  of  two  unequal  chords,  the  greater  subtends 
the  greater  minor  arc. 

164.  Corollary.  If  in  the  same  circle,  or  in  equal  circles, 
two  chords  are  equal,  the  angles  at  the  centre  luhich  they  subtend 
are  also  equal;  and  of  two  unequal  chords,  the  greater  subtends 
the  greater  angle. 

Proposition  V 

165.  The  straight  line  drawn  from  the  centre  of  a 
circle  perpendicular  to  a  chord  bisects  the  chord,  and 
if  produced,  bisects  the  arc  subtended  by  the  chord. 


Let  AB  be  any  chord  of  the  circle  AEB  whose  centre  is  0, 
and  OC  a  line  drawn  from  0  perpendicular  to  AB. 

It  is  required  to  prove  that  OC  bisects  the  chord  AB  at  C,  and 
if  produced,  bisects  the  arc  AB  at  D. 

Proof.     Join  OA  and  OB. 

As  OAP  and  OBC  are  identically  equal.         Why  ? 

Therefore  AC  =  BC. 

Also  ZAOD  =  ZBOD. 

Therefore  arc  AD  =  arc  BD.  (Art.  158.) 


106  ELEMENTARY  GEOMETRY  [Chap.  II 

166.  Corollary  I.  The  straight  line  drawn  fror)i  the  centime 
of  a  circle  to  the  mid-point  of  a  chord  is  perpendicidar  to  the 
chord. 

167.  Corollary  II.  Tlie  mid-points  of  a  system  of  parallel 
chords  all  lie  on  a  diameter  perpendicular  to  the  chords. 

For  the  diameter  drawn  to  the  mid-point  of  one  chord  is  per- 
pendicular to  that  chord  (Cor.  1),  and  therefore  to  every 
chord  of  the  system  (Art.  97).  Hence  this  diameter  passes 
through  the  mid-point  of  every  chord  of  the  system. 

168.  Corollary  III.  Tlie  straight  line  joining  the  centres 
of  tico  intersecting  circles  bisects  their  common  chord  at  right 
angles. 

Suggestion.  Join  the  mid-point  of  the  common  chord  with  the  centre 
of  one  circle.  This  line,  if  produced  backward,  will  pass  through  the 
centre  of  the  other  circle.     Why  ? 


169.  Corollary  IV.  The  diameter  which  bisects  a  chord 
also  bisects  the  angle  at  the  centre  subtended  by  the  chord. 

EXERCISES 

1.  If  two  chords  of  a  circle  intersect,  they  cannot  both  be  bisected  at 
their  common  point,  unless  that  point  is  the  centre. 

2.  Through  a  given  point  within  a  circle,  not  the  centre,  draw  a  chord 
which  is  bisected  at  that  point.    Is  there  more  than  one  such  chord  ? 

3.  Prove  that  the  arcs  of  a  circle  which  lie  between  two  parallel  chords 
are  equal. 

4.  The  straight  line  joining  the  mid-point  of  an  arc  of  a  circle  and  the 
mid -point  of  its  chord  is  perpendicular  to  the  chord,  and  will,  if  produced, 
pass  through  the  centre  of  the  circle. 

5.  The  diameter  which  bisects  an  arc  of  a  circle  also  bisects  its  chord 
at  right  angles. 

6.  Show  how  to  bisect  a  given  arc  of  a  circle. 

7.  If  a  quadrilateral  is  inscribed  in  a  circle,  the  perpendiculars  to  the 
sides  at  their  mid-points  meet  in  one  point. 


166-170] 


THE  CIRCLE 


107 


Proposition  VI 

170.  In  the  same  circle,  or  in  equal  circles,  equal 
chords  are  equidistant  from  the  centre ;  and  of  two  un- 
equal chords,  the  greater  is  nearer  to  the  centre  than 
the  less. 

^,  ^ ^^ 

C. 


First,  let  Oi  and  O^  be  the  centres  of  two  equal  circles, 
AiBi  and  A2B2  equal  chords  in  them. 

It  is  required  to  prove  that  the  perpendiculars  O^Ci  and 
O2C2  from  the  centre  upon  the  chords  are  equal. 

Proof.  The  angles  at  the  centre  A^O^B^^  and  ^202^83  are 
equal.     Give  reasons.     Also  Z  AiOiCi  =  Z.  A2O2C2.     Why  ? 

A  AiOiCi  is  identically  equal  to  A  A2O2C2.     Prove. 

Therefore  O^C^  =  O2O2. 

Next,  let  the  chord  A2D2  be  greater  than  the  chord  ^2-52- 

It  is  required  to  prove  that  the  perpendicular  O2C2  upon  the 
lesser  chord  is  greater  than  the  perpendicular  O2E2  upon  the 
greater  chord. 

Proof.  Since  the  chord  A2D2  is  greater  than  the  chord  ^2-^2? 
the  arc  A2D2  is  greater  than  the  arc  A2B2.  (Prop.  IV.) 

If  the  chords  be  so  placed  as  to  coincide  at  one  extremity 
and  lie  on  the  same  side  of  the  centre,  the  arc  A2D2  will  ex- 
tend beyond  the  arc  A2B2,  and  the  chord  ^2-^2  will  lie  on  the 
opposite  side  of  A2D2  from  the  centre.  Consequently,  the  per- 
pendicular O2C2  will  intersect  the  chord  A2D2  at  some  point  F2. 

Now  O2C2  is  greater  than  O2F2,  and  02i^2  is  greater  than  02^2- 
Why  ?     Therefore  O2C2  is  greater  than  02-E^2- 


108  ELEMENTARY  GEOMETRY  [Chap.  II 


Proposition  VII 

171.  In  the  same  circle,  or  in  equal  circles,  chords 
equidistant  from  the  centre  are  equal;  and  of  two 
chords  unequally  distant,  the  one  nearer  the  centre  is 
the  greater. 

What  is  the  relation  of  this  proposition  to  Proposition  VI  ? 

The  proof  is  left  to  the  pupil,  with  the  suggestion  that  the  indirect 
method  will  probably  be  easiest. 

Thus,  if  in  the  first  case,  the  chords  are  not  equal,  one  or  the  other  of 
them  must  be  nearer  the  centre.     Which  ?    Prop.  VI. 

In  the  second  case,  if  the  one  nearer  the  centre  is  not  the  greater,  what  ? 

172.  Corollary.  A  diameter  is  the  greatest  chord  that  can 
he  drawn  in  any  circle. 

This  follows  directly  from  the  above  proposition, 
or  it  may  be  shown  otherwise,  as  follows  :  — 

The  diameter  AB  equals  the  sum  of  the  radii     A 
AO  and  0(7,  which  is  greater  than  any  chord  ^C, 
not  a  diameter. 

EXERCISES 

1.  If  an  equilateral  triangle  is  inscribed  in  a  circle,  the  sides  are 
equidistant  from  the  centre. 

2.  If  two  chords  of  a  circle  which  intersect  make  equal  angles  with  the 
line  joining  their  common  point  to  the  centre,  show  that  the  chords  are 
equal. 

3.  A  chord  which  is  perpendicular  to  a  radius  is  less  than  any  other 
chord  through  their  point  of  intersection. 

4.  If  from  a  point  within  a  circle  more  than  two  equal  straight  lines 
can  be  drawn  to  the  circle,  that  point  must  be  the  centre. 

Suggestion.  —  Suppose  that  three  equal  lines  can  be  drawn  from  the 
point,  and  show  that  the  point  is  the  intersection  of  two  lines  upon  each 
of  which  the  centre  lies.     See  Art.  155. 

5.  Describe  two  concentric  circles  each  of  which  passes  through  two 
given  points,  the  first  through  A  and  B,  say,  and  the  second  through 
C  and  D. 


171-173]  THE  CIRCLE  109 

Proposition  VIII 

173.  Of  all  line-segments  which  can  he  drawn  to  a 
circle  from  a  point  within  it,  not  the  centre,  the  greatest 
is  that  which  passes  through  the  centre,  and  the  least  is 
that  ivhich,  if  produced  backward,  would  pass  through 
the  centre;  and  of  any  two  others,  the  greater  is  that 
which  makes  the  less  angle  with  the  greater  segment  of 
the  diameter  through  the  point. 


Let  S  be  any  point  within  a  circle  other  than  the  centre,  SA 
the  line  drawn  from  S  through  the  centre  to  the  circle,  SB  the 
line  from  S  to  the  circle  which,  if  produced  backward,  would 
pass  through  the  centre,  >iS(7  and  SD  any  other  straight  lines 
from  S  to  the  circle,  of  which  SC  makes  a  less  angle  with  the 
diameter  through  S  than  does  SD. 

It  is  required  to  prove  (1)  that  SA  is  the  greatest  line  from  S 
to  the  circle,  (2)  that  SB  is  the  least  line,  and  (3)  that  SC  is 
greater  than  SD. 

Proof.     Join  DC  and  OD. 
Eirst,  SA  equals  the  sum  of  ^0  and  00. 
But  the  sum  of  SO  and  00  is  greater  than  SO.     Why  ? 
Therefore  SA  is  greater  than  SO,  any  other   line  drawn 
from  S  to  the  circle. 

Next,  SB  equals  the  difference  between  SO  and  00. 
But  the  difference  between  SO  and  00  is  less  than  SO. 

(Ex.  3,  p.  48.) 


110  ELEMEN TABY  GEOMETR Y  [Chap.  II 

Therefore  SB  is  less  than  SC,  any  other  line  drawn  from  S 
to  the  circle. 

Lastly,  if  we  rotate  the  line  SG  about  S  into  the  position  SD, 
thereby  increasing  Z  ASC  (hypothesis),  we  at  the  same  time 
increase  Z  AOC  till  it  becomes  Z  AOD. 

Since  Z  AOC  is  less  than  Z  AOD,  the  supplementary  Z  /S'OC 
is  greater  than  the  supplementary  Z  SOD,  while  the  sides  con- 
taining these  angles  are  respectively  equal. 

Therefore  SG  is  greater  than  SD.  (Art.  79.) 

174.  Corollary.  From  any  point  within  a  circle  two  equal 
straight  lines  can  he  drawn  to  the  circle  ;  these  make  equal  angles 
with  the  diameter  through  the  point. 

EXERCISES 

1.  State  and  prove  a  theorem  for  a  point  without  a  circle  similar  to 
that  of  Proposition  VIII. 

2.  If  two  circles  intersect,  any  two  parallel  lines  drawn  through  the 
points  of  intersection  and  terminated  both  ways  by  the  circles  are  equal. 

3.  If  two  circles  intersect,  any  two  lines  drawn  through  one  point  of 
intersection,  making  equal  angles  with  the  line  of  centres  and  terminated 
both  ways  by  the  circles,  are  equal. 

4.  If  with  the  vertex  of  an  isosceles  triangle  for  centre  a  circle  is 
described  which  cuts  the  base  or  the  base  produced,  show  that  the  seg- 
ments of  the  base  line  intercepted  between  the  extremities  of  the  base 
and  the  circle  are  equal. 

175.  In  Proposition  VIII  we  saw  that  if  from  S,  any  point 
within  a  circle,  not  the  centre,  a  straight  line  SP  is  drawn  to 
the  circle,  and  the  line  rotated  about  S 
in  the  way  indicated  by  the  arrowhead, 
while  P  traverses  the  circle,  the  magni- 
tude SP  will  continuously  increase  till  P 
reaches  the  point  A.  After  this  it  will 
continuously  decrease  till  P  comes  to  coin- 
cide with  B,  when  it  will  again  begin  to 
increase. 


173-177]  THE  CIRCLE  .  Ill 

The  magnitude  8P  is  thus  a  variable  quantity,  varying  con- 
tinuously as  the  angle  ASP  varies  continuously.  It  has  a  maxi- 
mum value,  viz.  8 A,  and  a  minimum  value,  viz.  SB. 

If  the  point  S  is  chosen  on  the  circle,  SP  is  a  variable  chord 
whose  maximum  value  is  a  diameter  and  whose  minimum  value 
is  zero. 

Section  II 

ANGLES   INSCRIBED   IN  ARCS 

176.  Definition.  If  a  point  is  chosen  on  any  arc  of  a  circle, 
and  the  chords  are  drawn  from  it  to  the  extremities  of  the  arc, 
the  angle  between  these  chords  is  said  to  be  an  angle  in  the 
arc,  or  an  angle  inscribed  in  the  arc,  and  the  arc  is  said  to 
contain  the  angle. 

An  angle  in  an  arc  is  often  spoken  of  as  an  angle  in  the  seg- 
ment formed  by  the  arc  and  its  chord,  and  the  segment  then  is 
said  to  contain  the  angle. 

The  angle  ACB  is  said  to  be  inscribed  C 

in  the  arc  AEB,  or  i7i  the  segment  AEB  ;        X^     /\\ 
it  is  an  angle  at  a  point  of  the  circle  sub-     /         /       \      \ 
tended  by  the  ai'c  AFB,  or  subtended  by  the    I         /         \      ] 
chord  AB.  \   /  \ 

Sometimes    the    expression    ^an    angle     \Z. y 

stands   upon   an   arc'  is  used  instead   of        ^ ^ 

*is  subtended  by  an  arc'  ^ 

Exercise.  In  what  arc  is  ZABC  inscribed  ?  What  arc  does  it  sub- 
tend ?     What  chord  does  it  subtend  ? 

177.  Definition.  A  polygon  is  said  to  be  inscribed  in  a 
circle  when  its  vertices  lie  on  the  circle ;  and  the  circle  is  said 
to  be  circumscribed  about  the  polygon. 

Each  angle  of  an  inscribed  polygon  is  subtended  by  an  arc  of 
the  circle,  and  is  inscribed  in  the  conjugate  arc. 


112 


ELEMENTARY  GEOMETRY 


[Chap.  II 


Proposition  IX 

178.  An  angle  whose  vertex  is  on  a  circle  equals  half 
the  angle  at  the  centre  subtended  by  the  same  arc. 
G 


Fig.  2 

Let  AB  be  any  arc  of  a  circle,  ACB  an  angle  at  the  point  G 
of  the  circle,  and  AOB  the  angle  at  the  centre  subtended  by 
the  same  arc  AB. 

It  is  required  to  prove  that  the  angle  ACB  equals  half  the 
angle  AOB. 

Proof.     Draw  CO  and  produce  it  to  meet  the  circle  at  D. 

Z  OCA  =  Z  OAC.  Why  ? 

And  Z  AOD  equals  the  sum  of  A  0.40  and  OCA.   (Art.  100.) 

Therefore  Z  ACO  equals  half  of  ZAOD. 

Similarly  Z  jBOO  equals  half  of  Z  BOD.     Prove. 

Therefore  the  sum  (Fig.  1),  or  difference  (Fig.  2),  of  Z  ^00 
and  BCO  equals  half  the  sum,  or  difference,  of  A  AOD  and  BOD. 

That  is,  Z  ACB  equals  half  of  Z  AOB. 

o 


If  the  arc  subtending  the  angle  is  greater  than  a  semicircle, 
then  the  angle  at  the  centre  becomes  the  reflex  angle  AOB-, 


178-183]  THE  CIRCLE  113 

and  if  the  arc  is  a  semicircle,  the  angle  at  the  centre  becomes 
the  straight  angle  AOB.  The  same  method  of  proof  will  show- 
that  the  theorem  holds  in  these  cases  as  well  as  in  the  cases 
treated  in  the  proposition. 

179.  Corollary   I.     An   angle  in  a  semicircle   is   a  light 

angle. 

For  it  is  half  of  a  straight  angle.     Prove  also  by  joining  the 
centre  with  the  vertex  C. 

180.  Corollary  II.  A71  angle  in  an  arc  greater  than  a 
semicircle  is  acute,  and  an  angle  in  an  arc  less  than  a  semicircle 
is  obtuse. 

181.  Corollary  III.  All  angles  in  a  circle  subtended  by  the 
same  arc,  or  by  equal  arcs,  are  equal  ; 

Or  in  other  words,  all  angles  inscribed  in  the  same  segment, 
or  in  equal  segments,  of  a  circle  are  equal. 

For  they  are  all  equal  to  half  of  the  same  angle  at  the  centre  of 
the  circle. 

182.  Corollary  IV.  Equal  angles  on  the  same  base,  and 
on  the  same  side  of  it,  have  their  vertices  on  an  arc  of  a  circle 
of  which  the  given  base  is  the  chord.  

If  ACB  and  ADB  are  equal  angles  on  the  same 
side  of  AB,  the  circle  passing  through  A^  C,  and 
B^  must  also  pass  through  D.  For  if  not  it  will 
cut  AD  at  some  other  point  as  E.  Then  Z  AEB 
will  equal  ZACB.  (Cor.  III.) 

But  Z  AEB  is  greater  (or  less)  than  Z  ADB. 

(Art.  59.)  ^"^ ^ 

Therefore  Z.ACB  is  not  equal  to  /.ADB,  contrary  to  hypothesis. 

Therefore,  etc. 

183.  Corollary  V.  Tlie  circle  described  upon  the  hypote- 
nuse of  a  right  triangle  as  diameter  passes  through  the  vertex  of 
the  right  angle. 


114 


ELEMENTARY  GEOMETRY 


[Chap.  II 


Proposition  X 

184.   The  opposite  angles  of  any  convex  quadrilateral 
inscribed  in  a  circle  are  supplementary. 


Let  ABCD  be  any  convex  quadrilateral  inscribed  in  a  given 
circle. 

It  is  required  to  prove  that  the  angles  BAD  and  BCD,  also 
that  the  angles  ABC  and  ADC  are  supplementary. 

Proof.     Join  AC  and  BD. 

Z  BAC  =  Z  BDC ',  also  Z  DAC  =  Z  DBC       Why  ? 

Therefore  the  sum  of  ABAC  and  DAC  equals  the  sum  of 
ABDCsindDBC 

That  is,  Z  BAD  equals  the  sum  of  A  BDC  and  DBC 

But  the  sum  of  A  BDC  and  DBC  is  the  supplement  of 
Z  BCD.     Why  ? 

Therefore  Z  BAD  is  the  supplement  of  Z  BCD. 

That  the  angles  AlBC  and  ADC  are  supplementary  may  be 
proved  in  the  same  way,  or  it  follows  directly  from  the  fact 
that  the  sum  of  the  interior  angles  of  any  quadrilateral  is 
equal  to  four  right  angles. 

Conversely.  If  two  opposite  angles  of  a  convex  quadrilateral 
are  supplementary,  its  four  vertices  lie  on  a  circle. 

Definition.  Four  points  which  lie  on  the  same  circle  are 
said  to  be  concyclic 


184-186]  THE  CIRCLE  115 

Alternative  Proof  of  Proposition  X. 

To  prove  that  of  the  inscribed  quadrilateral  ABCD,  A  ABC 
and  ADC  are  supplementary. 

ZADC  equals  half  of  Z  AOC,  0  being 
the  centre  of  the  circle.  (Prop.  IX.) 

Z  ABC  equals  half  of  the  reflex  angle 
AOC 

The  sum  of  A  ABC  and  ADC  equals 
half  of  the  sum  of  the  two  angles  at  0, 
i.e.  half  of  four  right  angles. 

Therefore  the  sum  of  A  ABC  and  ADB 
equals  two  right  angles,  or  A  ABC  and  ADCq^yq  supplementary. 

EXERCISES 
1.   If  a  parallelogram  is  inscribed  in  a  circle  it  must  be  a  rectangle. 

Section  III 

SECANTS   AND  TANGENTS 

185.  Definition.  An  unlimited  straight  line  which  inter- 
sects a  circle  in  two  points  is  called  a  secant  of  the  circle. 

186.  If  a  secant  AB  of  any  circle  is  rotated  about  one  of  its 
points  of  intersection  A,  the  other  point  of  intersection  B  will 
move  along  the  circle  till  it  comes  j, 

to  coincide  with  A,  and  then  if  the 
rotation  is  continued  will  reappear 
jn  the  other  side  of  A. 

In  that  position  of  the  secant 
in  which  the  two  points  of  inter- 
section coincide  the  straight  line 
is  said  to  touch  the  circle,  or  to 
be  tangent  to  it,  and  the  common 
point  of  the  line  and  circle  is 
called  the  point  of  contact.  The  line  itself  is  called  a  tangent 
to  the  circle. 


116 


ELEMENTARY  GEOMETRY 


[Chap.  II 


Definition.  A  tangent  to  a  circle  is  a  straight  line  which 
meets  the  circle,  but  which  when  produced  does  not  cut  it. 

In  consequence  of  this  definition  every  point  of  a  tangent  to 
a  circle  lies  outside  the  circle  except  the  point  of  contact. 

Proposition  XI 

187.  The  perpendicular  to  a  diameter  of  a  circle  at 
one  extremity  is  a  tangent  to  the  circle;  and  any  other 
straight  line  through  that  extremity  will  cut  the  circle 
at  a  second  point. 

B 


Let  AKB  be  any  circle,  0  its  centre,  AB  a  diameter,  AF  a 
straight  line  perpendicular  to  the  diameter  at  the  point  A,  and 
AH  any  other  straight  line  through  A. 

It  is  required  to  prove,  (1),  that  AF  is  a  tangent  to  the  circle, 
and  (2),  that  AH  will  meet  the  circle  at  a  second  point  and 
consequently  is  not  a  tangent. 

Proof.     First,  choose  any  point  C  in  AF,  and  join  0(7. 

Since  AF  is  perpendicular  to  AB,  OC  is  greater  than  OA. 

Hence  C  lies  outside  the  circle. 

Since  AF  meets  the  circle  at  A,  while  every  other  point  of  it 
lies  outside  the  circle,  it  is  a  tangent  by  definition. 

Next,  draw  OD  perpendicular  to  AH,  meeting  AH  at  D. 

Then  OD  is  less  than  OA,  so  that  D  lies  within  the  circle, 
and  AD  produced  must  meet  the  circle  a  second  time.    (Art.  25.) 

Hence  AH  is  not  a  tangent  to  the  circle. 


186-192]  THE  CIRCLE  117 

188.  Corollary  I.  At  any  point  of  a  circle  there  can  he 
drawn  one  and  only  one  tangent. 

For  at  the  extremity  of  the  diameter  through  the  point  there  can 
be  drawn  one  and  only  one  perpendicular  to  this  diameter. 

189.  Corollary  II.  Any  tangent  to  a  circle  is  perpendicular 
to  the  radius  drawn  to  the  point  of  contact. 

For,  if  not,  it  must  cut  the  circle  at  a  second  point. 

190.  Corollary  III.  The  centre  of  a  circle  lies  on  the 
perpendicular  to  a  tangent  drawii  from  the  point  of  contact. 

For  the  straight  line  joining  the  centre  to  the  point  of  contact 
is  perpendicular  to  the  tangent,  and  there  can  be  only  one 
perpendicular  to  the  tangent  drawn  from  the  point  of  contact. 

191.  Corollary  IV.  The  straight  line  drawn  from  the 
centre  of  a  circle  perpendicular  to  a  tangent  meets  it  at  the 
point  of  contact. 

192.  Corollary  V.  The  tangent  to  a  circle  at  the  mid-point 
of  any  arc  is  parallel  to  the  chord  of  the  arc. 

For  it  is  perpendicular  to  the  radius  drawn  to  its  point  of  con- 
tact, and  the  chord  is  also  perpendicular  to  this  radius. 

(Ex.  5.  p.  106.) 

EXERCISES 

1.  Draw  a  tangent  to  a  circle  from  a  given  point  on  the  circle. 

2.  Draw  a  tangent  to  a  circle  which  shall  be  (1)  parallel  to  a  given 
straight  line;  (2)  perpendicular  to  a  given  straight  line.  How  many 
such  tangents  can  be  drawn  ? 

3.  Draw  a  tangent  to  a  circle  which  shall  make  a  given  angle  with  a 
given  straight  line.     How  many  such  tangents  can  be  drawn  ? 

4.  A  straight  line  will  intersect,  touch,  or  lie  wholly  outside  of  a  given 
circle  according  as  its  distance  from  the  centre  is  less  than,  equal  to,  or 
greater  than  a  radius  of  the  circle. 


118  ELEMENTARY  GEOMETRY  [Chap.  II 


Proposition  XII 

193.   To  draw  a  tangent  to  a  circle  from  a  given  point 
without  it. 


Let  BCD  be  any  given  circle  whose  centre  is  0,  and  let  A  be 

any  point  without  it. 

It  is  required  to  draw  from  A  a  straight  line  tangent  to  the 
given  circle. 

Construction.     Join  AO,  and  upon  ^0  as  diameter  describe  a 
circle  intersecting  the  given  circle  at  B  and  C.     Join  AB. 
Then  AB  is  a  tangent  to  the  given  circle. 
Proof.     Join  OB.     Z  OBA  is  a  right  angle.     Why  ? 
Therefore  AB  is  tangent  to  the  given  circle  at  B.    (Prop.  XI.) 
Similarly  ^C  is  tangent  to  the  given  circle  at  G. 

194.  Corollary.  The  segments  of  two  tangents  to  a  circle 
which  lie  between  their  point  of  intersection  and  their  points  of 
contact  are  equal,  and  make  equal  angles  with  the  line  joining 
their  intersection  to  the  cent7'e  of  the  circle. 

Suggestion.  In  the  above  diagram  prove  AB  equal  to  AC,  and 
ZBAO  equal  to  Z  CAO. 

Sometimes,  when  no  misunderstanding  is  likely  to  arise,  the 
segment  of  a  tangent  lying  between  the  point  of  contact  and 
the  point  A  from  which  it  is  drawn  will  be  spoken  of  as  the 
tangent  from  A.  In  that  case  the  above  corollary  may  be 
stated  as  follows : 


193-196]  THE  CIRCLE  119 

Corollary.  The  two  tmigents  to  a  circle  drawn  from  any 
outside  point  are  equal,  and  the  straight  line  from  the  point  to 
the  centre  of  the  circle  bisects  the  angle  between  them. 

Definition.  The  straight  line  joining  the  points  of  contact 
of  two  tangents  is  called  their  chord  of  contact. 

195.  From  the  above  construction  two  tangents  can  be  drawn  to  a 
circle  from  any  outside  point,  and  the  question  immediately  arises,  can 
more  than  two  tangents  be  drawn  from  the  same  point  ? 

Suppose  in  the  above  diagram  another  straight  line  be  drawn  through 
A  meeting  the  given  circle  at  a  point  E,  different  from  B  or  C. 

Then  if  AE  is  also  tangent  to  the  circle,  Z  OEA  is  a  right  angle,  and 
the  circle  whose  diameter  is  ^0  would  pass  through  E  (Art.  183), 
and  would  therefore  intersect  the  given  circle  at  E. 

But  this  is  impossible  since  this  circle  already  intersects  the  given  circle 
at  two  points,  B  and  C.  (Art.  151.) 

Therefore  no  third  tangent  can  be  drawn  through  a  given  point. 

Hence  the  theorem, 

Two  tangents  and  only  two  can  be  drawn  to  a  circle  from  an 
outside  point. 

196.  A  polygon  is  said  to  be  circumscribed  about  a  circle 
when  its  sides  are  all  tangent  to  the  circle;  and  the  circle  is 
said  to  be  inscribed  in  the  polygon. 

If  a  triangle  is  circumscribed  about  a  circle,  the  bisector 
of  any  of  its  angles  passes  through  the  centre  of  the  circle 
(Art.  194). 

Hence  to  solve  the  problem : 

To  inscribe  a  circle  in  a  given  triangle, 

you  need  only  to  bisect  two  of  the  angles  of  the  triangle,  and 
the  point  of  intersection  of  these  bisectors  will  be  the  centre  of 
th^  required  circle.  For  each  bisector  is  the  locus  of  points 
equidistant  from  two  sides  of  the  triangle,  hence  their  point 
of  intersection  is  equidistant  from  all  three  sides. 


120  ELEMENTARY  GEOMETRY  [Chap.  II 


EXERCISES 

1.  The  angle  between  any  two  tangents  to  a  circle  is  supplementary  to 
the  angle  between  the  radii  drawn  to  their  points  of  contact. 

2.  If  a  quadrilateral  is  circumscribed  about  a  circle,  the  sum  of  two 
opposite  sides  is  equal  to  the  sum  of  the  other  two  opposite  sides. 

3.  If  a  parallelogram  is  circumscribed  about  a  circle  it  must  be  a 
rhombus  and  its  diagonals  must  pass  through  the  centre. 

4.  The  diameter  through  the  intersection  of  two  tangents  bisects  their 
chord  of  contact  at  right  angles. 

5.  Prove  that  the  chord  which  joins  the  points  of  contact  of  two 
parallel  tangents  to  a  circle  is  a  diameter. 

6.  If  a  circle  is  described  upon  one  side  of  an  isosceles  triangle  as 
diameter,  show  that  it  will  pass  through  the  mid-point  of  the  base. 

7.  Construct  a  right  triangle  having  given  the  hypotenuse  and  one 
side. 

8.  If  ABC  is  a  triangle  whose  vertices  lie  on  a  circle  and  AD  bisects 
the  angle  at  A  meeting  the  circle  again  at  Z>,  show  that  the  diameter 
through  D  bisects  J5(7  at  right  angles. 

9.  If  the  opposite  sides  AB  and  CD  of  a  quadrilateral  ABCD  inscribed 
in  a  circle  are  produced  to  meet  in  E^  the  triangles  AEC  and  DEB  are 
equiangular,  as  are  also  the  triangles  AED  and  CEB. 

10.  Divide  a  circle  into  two  arcs  such  that  the  angle  contained  by  one 
shall  be  twice  ^the  angle  contained  by  the  other. 

11.  Chords  of  a  circle  whose  centre  is  0  pass  through  a  fixed  point  P; 
show  that  the  locus  of  the  mid-points  of  the  chord  is  a  circle  whose 
diameter  is  OP. 

12.  Two  circles  intersect  at  the  points  A  and  B,  and  a  straight  line 
PA  Q  is  drawn  through  A  cutting  the  circles  at  P  and  Q,  respectively. 
Show  that  if  the  circles  are  equal  the  chords  PB  and  QB  are  equal,  and 
conversely,  if  PB  and  QB  are  equal  the  circles  must  be  equal. 

13.  If  a  circle  is  inscribed  in  a  right  triangle,  the  sum  of  its  diameter 
and  the  hypotenuse  of  the  triangle  is  equal  to  the  sum  of  the  other  two 
sides. 


196-198]  THE  CIRCLE  121 


Proposition  XIII 

197.  The  angle  formed  by  a  tangent  to  a  circle  and  a 
chord  through  its  point  of  contact  equals  half  the  angle 
at  the  centre,  subtended  by  the  intercepted  arc. 


Let  0  be  the  centre  of  the  given  circle,  PQ  a  tangent  at  A, 
and  AB  a  chord. 

It  is  required  to  prove  that  the  angle  QAB  equals  half  the 
angle  AOB  at  the  centre,  subtended  by  the  arc  AB. 

Proof.  Draw  the  diameter  AO,  meeting  the  circle  a  second 
time  at  C,  and  join  BO  and  BC. 

Z  BAQ  is  the  complement  of  Z  BAC.     Why  ? 

/-ACB  is  also  the  complement  of  Z  BAC.     Why  ? 

Therefore  Z  BAQ  equals  A  ACB. 

But  Z  ACB  equals  half  of  Z  AOB.  (Prop.  IX.) 

Therefore  Z  BAQ  equals  half  of  ZAOB. 

Show  in  a  similar  manner,  or  otherwise,  that  the  angle  FAB  equals 
half  of  the  reflex  angle  AOB. 

198.  Corollary.  The  angle  between  a  tangent  to  a  circle 
and  a  chord  through  its  point  of  contact  is  equal  to  any  angle 
inscribed  in  the  segment  on  the  opposite  side  of  the  chord. 

By  'the  intercepted  arc'  is  meant  the  arc  falling  within  the  angle. 
Thus,  if  we  have  in  mind  Z  QAB  as  the  angle  between  the  tangent  and 


122  ELEMENTARY  GEOMETRY  [Chap.  II 

the  chord,  the  intercepted  arc  is  the  minor  arc  AB ;  but  if  we  think  of 
Z  PAB  as  the  angle  between  the  tangent  and  the  chord,  the  intercepted 
arc  is  the  major  arc  AB. 

The  corollary  states  that  Z  QAB  equals  any  angle  inscribed  in  the 
segment  ACB^  and  that  ZPAB  equals  any  angle  inscribed  in  the  seg- 
ment ADB. 


Proposition  XIV 

199.   On  a  given  line-segment  as  chord,  to  construct  an 
arc  of  a  circle  which  shall  contain  a  given  angle. 


Let  AB  be  the  given  line-segment,  and  C  the  given  angle. 

It  is  required  to  construct  on  AB  an  arc  of  a  circle  which 
shall  contain  an  angle  equal  to  the  angle  C. 

Construction.  From  one  extremity  of  the  given  line-segment 
draw  a  straight  line  AD  making  Z  BAD  equal  to  the  given 
angle  C 

AB  is  to  be  the  chord  of  the  required  arc,  and  if  AD  were  a 
tangent  the  given  angle  would  be  equal  to  the  angle  between 
a  tangent  and  chord  of  the  required  circle. 

The  problem  then  becomes :  To  describe  a  circle  tangent  to 
AD  at  A  and  passing  through  B. 

Find  the  centre  of  such  a  circle  and  complete  the  problem, 
giving  references  to  all  theorems  employed. 


198-201]  THE  CIRCLE  123 

Proposition   XV 

200.  The  angle  formed  hy  two  secants  which  intersect 
within  a  circle  is  equal  to  the  sum  of  the  two  angles 
whose  vertices  are  on  the  circle,  and  which  are  sub- 
tended hy  the  intercepted  arcs. 


Let  AB  and  CD  be  any  two  secants  of  a  circle  intersect- 
ing at  P  within  the  circle. 

It  is  required  to  prove  that  the  angle  APC  is  equal  to  the 
sum  of  the  angles  at  points  of  the  circle,  subtended  by  the 
arcs  AC  and  BD. 

Proof.  Join  EC  Then  Z  APC  equals  the  sum  of  Zs  PEG 
and  ECP.     Why  ? 

But  Z  PECy  or  Z  AEC,  has  its  vertex  on  the  circle  and  is 
subtended  by  the  arc  AC. 

And  Z  ECP,  or  Z  BCD,  is  subtended  by  the  arc  ED. 

Therefore,  etc. 

In  just  the  same  way  it  can  be  shown  that  the  supplementary 
angle  APD  equals  the  sum  of  the  angles  subtended  by  the 
arcs  AD  and  EC. 

201.  If  the  point  of  intersection  P  lies  on  the  circle  one  of  the  inter- 
cepted arcs  vanishes,  but  the  theorem  as 
stated  for  an  internal  point  is  still  true. 
For  ZAPC,  or  its  vertical  angle,  is  sub- 
tended by  the  arc  BD,  while  ZAPD 
equals  the  sum  of  the  angles  whose  vertices 
are  on  the  circle  and  which  are  subtended 
by  the  arcs  BP  and  PD.  This  wUl  readily 
be  seen  if  you  join  BD, 


124  ELEMENTARY  GEOMETRY  [Chap.  II 


Proposition  XVI 

202.  The  angle  formed  by  two  secants  which  intersect 
without  a  circle  is  equal  to  the  difference  between  the 
two  angles  whose  vertices  are  on  the  circle  and  which 
are  subtended  by  the  intercepted  arcs. 


Let  AB  and  CD  be  any  two  secants  of  a  circle,  intersecting 
at  P,  without  the  circle. 

It  is  required  to  prove  that  the  angle  APC  is  equal  to  the 
difference  between  the  angles  at  points  of  the  circle,  subtended 
by  the  arcs  AC  and  BD. 

Proof.  Join  BC.  Then  Z  APC  equals  the  difference  be- 
tween Zs  ABC  and  BCD.  (Art.  100.) 

But    Z.ABC  is  subtended  by  the  arc  AC. 

And  Z  BCD  is  subtended  by  the  arc  BD. 

Therefore,  etc. 

In  this  case  the  supplementary  angle  APx  equals  the  sum, 
not  the  difference,  of  Zs  ABD  and  BDC,  or  the  sum  of  the 
angles  subtended  by  the  arcs  AD  and  BC. 

EXERCISES 

1.  Construct  a  triangle  having  given  the  base,  the  vertical  angle,  and 
one  side.     What  is  the  limitation  on  the  length  of  this  side  ? 

2.  Construct  a  triangle  having  given  the  base,  the  vertical  angle, 
and  the  altitude.    What  is  the  limitation  upon  the  altitude  ? 

3.  Construct  a  triangle  having  given  the  base,  the  vertical  angle,  and 
the  sum  of  the  sides. 


202-204] 


THE   CIRCLE 


125 


203.  Suppose  one  of  the 
secants  AB  becomes  a  tangent 
at  A^  the  two  intersection 
points  A  and  B  thus  coinciding 
at  A.  Then  ZAPC  between 
the  tangent  and  the  secant 
equals  the  difference  between 
Z  ACD  and  Z  CAP.  But 
/LACD  is  subtended  by  the  in- 
tercepted arc  AD^  or  BD  since  A  and  B  coincide  ;  and  Z  CAP  equals 
an  angle  subtended  by  the  intercepted  arc  AC.    (Prop.  XIII.) 

Hence,  just  as  in  the  case  of  two  secants,  we  have, 

Theorem.  The  angle  between  a  tangent  and  a  secant  equals  the  differ- 
ence between  the  angles  whose  vertices  are  on  the  circle  and  which  are 
subtended  by  the  intercepted  arcs. 

The  supplementary  angle  APx  equals  the  sum  of  Zs  PAD  and  PDA. 

But  Z  PAD  equals  an  angle  subtended  by  the  arc  ACD,  while  Z  PDA 
is  subtended  by  the  arc  AC. 

That  is,  the  supplementary  angle  APx  equals  the  sum  of  the  angles 
subtended  by  the  arcs  AD 

and  AC  (or  BC),  just  as  in         \  ,  w 

the  case  of  the  two  secants. 

Suppose,  finally,  that  both 
secants  become  tangents,  the 
intersection  points  A  and  B 
coinciding  at  A,  while  C  and 
D  coincide  at  C. 

Then  ZAPC  equals  the 
difference  between  Zs  ACz 
and  PAC,  i.e.  equals  the 
difference  between  the  angles 
subtended  by  the  arcs  AKC 
and  ALC.  And  ZAPy  equals  the  sum  of  ZsP^C  and  PCA,  i.e.  equals 
twice  the  angle  subtended  by  the  arc  ALC. 


Principle  of  Continuity 

204.  The  preceding  discussion  of  the  different  phases  of  Propositions 
XV  and  XVI  leads  us  to  the  consideration  of  a  principle  which  is  gen- 
erally recognized  in  geometry  as  fundamental.  We  shall  take  up  in  the 
first  place  some  very  simple  illustrations  of  the  principle  and  then  proceed 


126  ELEMENTARY   GEOMETRY  [Chap.  II 

to  more  complicated  ones,  such  as  are  involved  in  the  theorems  just  con- 
sidered. 

1.    Suppose  we  have  given  any  line-segment  AB  marked  out  on  an 
unlimited  straight  line  m.    If  we  choose  a  point  C  on  this  line  between  A 


and  5,  we  say  that  the  line-segment  AB  equals  the  sum  of  the  line- 
segments  AC  and  CB. 

Or,  expressed  algebraically, 

AB  =  AC-{-CB. 

This  is  true  no  matter  where  the  point  C  is  chosen  between  A  and  B^ 
or  if  C  coincides  with  either  A  or  B^  for  in  the  latter  case  one  of  the  seg- 
ments becomes  zero  and  the  other  equals  AB. 

If,  however,  G  is  chosen  on  the  line  m,  not  between  A  and  B,  we 


are  accustomed  to  say  that  the  line-segment  AB  equals  the  difference 
between  the  line-segments  ^C  and  BC^ 
Or,  expressed  algebraically, 

AB  =  AC-BC. 

In  order  to  avoid  the  necessity  of  varying  the  statement  for  different 
positions  of  the  point  O,  it  is  convenient  to  think  of  a  line-segment  AB  as 
having  a  definite  sense,  i.e.  as  though  it  started  at  A  and  ran  to  B.  The 
line-segment  BA  would  then  run  in  the  opposite  sense. 

If  we  attach  the  algebraic  term  positive  to  the  line-segment  AB^  then 
the  line-segment  BA  should  be  thought  of  as  negative.,  since  positive  and 
negative  always  indicate  opposites  in  sense. 

Thus  the  line-segment  AB  =  —  the  line-segment  BA, 

and  "  "  BA  =  -    "  "  AB. 

So  for  any  line-segment  whatever  PQ  =  —  QP. 

With  this  understanding,  the  expression  AC  —  BC  is  just  the  same  as 
AC+  CB. 

Hence,  if  we  attach  to  a  line-segment  a  distinguishing  sign,  positive  or 
negative,  according  as  it  is  traversed  in  the  one  seiise  or  in  the  other,  no 
matter  where  the  point  C  may  be  chosen  upon  the  unlimited  straight  line, 
wheiher  between  A  and  B  or  not,  it  is  always  true  that 

AB  =  AC-{-CB. 


204]  THE  CIRCLE  127 

So  also,  no  matter  how  the  points  M^  JV,  P,  may  be  placed  upon  a 
straight  line, 

MN-\-NP=MP, 

or  expressed  in  words,  the  result  of  going  from  M  to  N  and  then  from  N 
to  P  is  the  same  as  going  from  M  to  P. 

So  again  for  any  four  points  P,  ^,  P,  iS',  upon  a  straight  line, 

Pq+  QE-^BS=FS, 

or  PB  +  IiS+  SQ  =  PQ. 

Many  theorems  in  geometry  consist  simply  of  a  statement  of  some 
relation  among  the  parts  of  a  geometrical  figure,  and  the  principle  to 
which  we  have  referred  affirms  that  this  relation,  once  true  and  properly 
interpreted,  remains  true  when  the  figure  changes  continuously  from  one 
form  to  another,  subject  to  the  conditions  under  which  it  was  first  de- 
scribed. For  this  reason  the  principle  is  called  the  principle  of  con- 
tinuity. 

Thus  in  the  case  just  considered 

AB  =  AC-^  CB 

when  C  lies  on  the  line  m  between  A  and  B.  The  principle  of  continuity 
affirms  that  this  relation  remains  true  when  C  moves  anywhere  along  the 
line  ?7i,  if  the  signs  of  the  segments  be  taken  into  account. 

2.  In  Proposition  X  we  proved  that  if  AQBP  is  a  convex  quadri- 
lateral inscribed  in  a  circle,  the  angles  BQA  and  BPA  are  supplementary, 
and  it  follows  immediately  that  Z  PPT  equals  ZBQA. 

Now  suppose  in  the  figure  that  the  points  A,  Q,  P,  remain  fixed  while 
P  moves  along  the  circle  toward  A,  and  finally  coincides  with  A.  During 
this  motion  ZBPA  remains  constant  (Prop.  IX,  Cor.  Ill),  as  does  also 
Z.BPT.  When  P  comes  to  coincide  with  A^  AP  becomes  the  tangent 
at  A,  and  BP  takes  the  position  BA  and  Z  BPT  ijf 

falls  into  the  position  of  ZBAT.  / 

But  ZBPT^ZBQA.  \^         I 

Therefore   ZBATz=  ZB^A.  \      I  J:^^—^^ 

This  last  statement,  however,  is  nothing  else  l^^^"'''^     _     1^  r 

than  the  theorem  of  Proposition  XIII.    So,  by  /^^  ^y\ 

letting  the  secant  A  T  vary  continuously  till  it  I  \\  ^^^  /    I 

takes  the  position  of  a  tangent,  we  pass  from  \  \     \  ^''''    /       / 
the  theorem  of  Proposition  X  to  that  of  Propo-  \\  •''  \     /      / 

sition  XIII.     In  other  words,  Proposition  XIII  P^'^^-J^S/^'-'^ 

is  merely  a  special  case  of  Proposition  X.  Q 


128  ELEMENTARY  GEOMETRY  [Chap.  II 

If  the  rotation  of  the  secant  A  T  is  continued  beyond  the  position  of  a 
tangent,  Zs^PTand  BQA  become  angles  in  the  same  segment. 

During  the  process  our  figure  has  changed  its  form  continuously,  and 
the  theorem  with  which  we  started  has  developed  forms  which  were  at 
first  considered  to  be  different,  but  which  we  now  see  to  be  merely  differ- 
ent phases  of  the  same  theorem. 

3.  Now  turn  to  Proposition  XV.  The  theorem  states  that  ZBPD 
equals  the  sum  of  the  angles  whose  vertices  are  on  the  circle  and  w^hich 
are  subtended  by  the  arcs  BD  and  AC,  and  it  was  shown  in  connection 
with  this  proposition  that  Z  APD  equals  the  sum  of  the  angles  subtended 
in  the  same  way  by  the  arcs 
AD  and  BC.  ^ 

It  should  be  noticed  that       ^^^^*v^  /'""         "\ 

the  arcs  mentioned  above  have  -f/^<;;^~- — C//  \ 

been  named  in  each  case,  start-  _l!!^^:^  _  V"~ — ::^~- ■   \ 

ing  from  the  points  A  and  B;  '\^^^''^^       ^^^-^^l 

also  that  if  we  think  of  an  in-  \  ^..-----^pCr  / 

itial  and  a  terminal  point  of  ^-^^"^^x  ^"^^Nc 

an  arc,  as  we  did  of  a  line-  \^^^    _^^''^^'^*^ 

segment,    then   AC  and   BD 
run  round  the  circle  in  the  same  sense  ;  so  also  do  AD  and  BG. 

Suppose  now  we  make  an  agreement  as  to  the  signs  to  be  attached  to 
arcs,  similar  to  that  which  was  made  upon  line-segments,  viz.,  arcs 
which  run  round  the  circle  in  the  same  sense  shall  have  the  same  sign, 
while  arcs  which  run  round  the  circle  in  opposite  senses  shall  have  oppo- 
site signs. 

So  also  angles  subtended  by  arcs  of  the  same  sign  shall  be  of  the  same 
sign,  while  angles  subtended  by  arcs  of  opposite  signs  shall  be  of  opposite 
signs. 

In  the  figure  before  us  let  us  hold  the  points  A,  B,  D,  fixed  and  rotate 
the  line  DC  about  D  in  the  sense  indicated  by  the  arrowhead.  Then  C 
will  move  along  the  circle  toward  A  till  it  coincides  with  A,  and  will 
then  pass  to  the  other  side.  The  intersection  point  P  of  the  two  secants 
must  move  out  along  BA  till  it  likewise  coincides  with  A  and  will  then  go 
beyond. 

To  start  with  we  have  these  two  relations  from  Proposition  XV. 

Z  DPB  =  angle  subtended  by  arc  DB  +  angle  subtended  by  arc  CA. 
Z  DPA  =  angle  subtended  by  arc  DA  +  angle  subtended  by  arc  CB. 

When,  by  the  rotation  of  DC,  C  comes  to  coincide  with  A,  the  arc  CA 
vanishes,  but  the  two  relations  are  easily  seen  to  remain  true. 


204]  THE  CIRCLE  129 

If  G  passes  beyond  A  into  the  position  Ci,  say,  so  that  P  goes  outside 
the  circle,  the  arcs  CiA  and  DB  have  opposite  signs,  and  the  angles 
subtending  them  must  also  be  considered  as  having  opposite  signs.  The 
algebraic  sum  on  the  right  of  the  first  relation  then  becomes  the  numeri- 
cal difference  of  the  angles,  and  the  relation  expressed  becomes  the  same 
as  that  stated  in  Proposition  XVI. 

As  P  passes  outside  the  circle,  the  arcs  CB  and  DA  overlap,  but 
remain  of  the  same  sign.  So  the  second  relation  is  unaltered  for  Propo- 
sition XVI. 

The  figure  is  capable  of  still  further  variation  which  would  serve  to 
develop  the  other  special  cases  given  under  Proposition  XVI.  For 
instance,  when  the  point  P  is  outside,  we  might  hold  P  fixed,  and  rotate 
PD  till  it  becomes  a  tangent.  The  arcs  AC  and  BD  still  have  opposite 
signs,  while  AD  and  BC  have  the  same  signs,  etc. 

Thus  by  making  application  of  the  principle  of  continuity,  and  by 
suitable  conventions  as  to  sign,  Proposition  XVI  and  the  special  cases 
which  were  considered  under  it,  are  shown  to  be  simply  modifications  of 
Proposition  XV.  And  so  it  frequently  happens  that  propositions  which 
are  at  first  glance  very  different,  can,  by  the  application  of  the  principle 
of  continuity,  be  harmonized  under  one  general  statement. 

EXERCISES 

1.  If  from  the  point  of  contact  of  a  tangent  to  a  circle,  a  chord  is  drawn, 
the  perpendiculars  from  the  mid-point  of  either  of  the  arcs  so  formed  to 
the  chord  and  the  tangent  are  equal. 

2.  ABC  is  a  triangle  inscribed  in  a  circle,  and  from  any  point  D  inBC 
a  straight  line  DE  is  drawn  parallel  to  CA^  meeting  the  tangent  at  A  in  E. 
Show  that  a  circle  may  be  described  through  the  four  points  J.,  E,  B,  D. 

3.  Two  circles  intersect  at  A  and  J5,  and  through  A  two  straight  lines* 
PAQ  and  BAS  are  drawn  cutting  one  circle  in  P  and  B,  respectively, 
and  the  other  in  Q  and  S.  Show  that  if  the  chord  PB  is  equal  to  the 
chord  QS,  the  circles  must  be  equal ;  and  conversely,  that  if  the  circles 
are  equal  the  chords  PB  and  QS  will  be  equal. 

4.  Prove  that  the  line  bisecting  any  interior  angle  of  a  quadrilateral 
inscribed  in  a  circle  and  the  line  bisecting  the  opposite  exterior  angle 
intersect  on  the  circle. 

5.  Prove  that  the  straight  lines  which  bisect  the  vertical  angles  of  all 
triangles  having  the  same  base  and  equal  vertical  angles,  have  one  point 
in  common. 


130 


ELEMENTARY  GEOMETRY 


[Chap.  11 


Section  IV 

CIRCLES  IN  CONTACT 

205.  Definition.  Two  circles  are  said  to  touch  each  other, 
or  to  be  in  contact,  when  they  have  a  common  tangent  at  a  com- 
mon point. 

This  common  point  is  called  the  point  of  contact. 


Proposition  XVII 

206.  When  two  circles  touch  each  other  their  point  of 
contact  and  their  tiuo  centres  lie  in  a  straight  line. 

For  the  perpendicular  to  their  common  tangent  at  the  point 
of  contact  must  pass  through  the  centre  of  each  circle. 

(Prop.  XI.  Cor.  III.) 

Proposition  XVIII 

207.  Two  circles  which  touch  each  other  can  have  no 
other  point  in  cominon  than  the  point  of  contaxit. 


O  Oi 


O  ^--^^. 


By  the  preceding  theorem  the  point  of  contact  and  the  two 
centres  are  collinear. 

The  two  centres  cannot  coincide,  for  then  the  circles  would 
be  identical,  having  a  common  centre  and  a  common  point. 

Two  cases  must  then  be  considered,  according  as  the  centres 
lie  (1)  on  the  same  side,  (2)  on  opposite  sides  of  the  point  of 
contact. 

Let  0  and  Oj  be  the  centres  of  the  two  circles,  and  let  A  be 
their  point  of  contact. 


205-208] 


THE  CIRCLE 


131 


If  possible,  let  P  be  another  point  common  to  the  two  circles. 

Then  OA  =  OP,  and  O^A  =  0,P. 

Hence  in  either  of  the  two  possible  cases, 
one  side  of  the  triangle  OPOi  must  equal 
the  sum  of  the  other  two  sides,  which  is 
impossible. 

Therefore,  two  circles  tangent  at  A  can 
have  no  other  common  point. 

When  the  two  centres  lie  on  the  same 
side  of  the  common  tangent,  the  circles 
are  said  to  have  internal  contact,  or  to 
touch  each  other  internally ;  when  the 
centres  lie  on  opposite  sides  the  circles  have  external  contact, 
or  touch  each  other  externally. 

In  the  first  case  one  circle  lies  inside  of  the  other,  in  the 
second  case  each  lies  outside  of  the  other. 


208.  Corollary.  If  two  circles  have  a  common  point  on  their 
line  of  centres,  they  must  have  a  common  tangent  at  that  point. 

For  the  tangent  to  either  circle  at  that  point  is  perpendicular  to  the 
line  of  centres  and  the  two  tangents  must  therefore  coincide.  That  is 
the  two  circles  have  a  common  tangent  at  that  point. 


EXERCISES 

1.  Two  circles  touch  each  other  externally  or  internally  at  a  point,  and 
through  that  point  a  straight  line  is  drawn  to  cut  the  two  circles.  If  the 
points  of  intersection  be  joined  to  the  respective  centres,  the  two  straight 
lines  so  drawn  will  be  parallel. 

2.  If  the  distance  between  the  centres  of  two  circles  is  equal  to  the  sum 
of  their  radii,  the  two  circles  must  touch  each  other  externally. 

3.  If  the  distance  between  the  centres  of  two  circles  is  equal  to  the 
difference  of  their  radii,  the  two  circles  must  touch  each  other  internally. 

4.  Describe  a  circle  passing  through  a  given  point  and  touching  a  given 
circle  at  a  given  point. 


132  ELEMENTARY  GEOMETRY  [Chap.  II 


Proposition  XIX 

209.  To  draw  a  cormnon  tangent  to  two  given  circles, 
each  of  which  lies  wholly  outside  of  the  other. 


Let  C  and  Oj  be  the  given  circles,  0  and  Oi  their  centres, 
and  suppose  the  radius  of  C  greater  than  the  radius  of  Ci- 

It  is  required  to  draw  a  straight  line  which  shall  touch  both 
of  these  circles. 

Construction.  With  centre  0  and  radius  equal  to  the  differ- 
ence between  the  radii  of  C  and  (7i  describe  a  circle  Gz- 

From  Oi  draw  a  tangent  to  C^,  the  point  of  contact  being  Tg. 
[Notice  that  two  such  tangents  can  be  drawn.] 

Join  07^2  and  produce  to  meet  C  at  T. 

From  Oi  draw  a  radius  of  C^,  viz.  OiT^,  parallel  to  OT,  and 
on  the  same  side  of  the  line  of  centres. 

Draw  the  straight  line  TT^,  which  shall  be  the  required 
common  tangent. 

Proof.  The  quadrilateral  TT^O^T^  is  a  parallelogram.  Show 
why. 

The  angle  O^T^T  is  a  right  angle.     Why  ? 

Therefore  the  straight  line  TT^  is  perpendicular  to  a  radius 
of  each  of  the  given  circles  at  points  T  and  T^  of  the  circles. 

It  is  therefore  tangent  to  both  circles. 

EXERCISES 

1.  Show  that  by  drawing  the  other  tangent  from  Oi  to  the  circle  C2, 
a  second  common  tangent  to  the  two  given  circles  could  be  found. 


209]  THE  CIRCLE  133 

2.  Show  that  by  drawing  a  circle  Cs  with  centre  Ox  instead  of  0, 
and  with  radius  equal  to  the  sum  of  the  radii  of  the  given  circles,  and  also 
by  drawing  the  parallel  radii  on  opposite  sides  of  the  line  of  centres,  two 
other  common  tangents  to  the  given  circles  could  be  found. 

The  two  common  tangents  first  drawn  are  called  the  direct  common 
tangents,  the  other  two  the  transverse  common  tangents. 

3.  Show  that  when  the  two  given  circles  are  equal  the  direct  common 
tangents  are  parallel.     How  does  the  construction  differ  in  this  case  ? 

4.  Show  that  when  the  two  given  circles  are  in  contact,  the  centre 
0  lies  upon  the  circle  Cz  (as  drawn  to  obtain  the  transverse  tangents)  ; 
and  that  when  the  given  circles  intersect,  the  centre  O  lies  within  the 
circle  O3. 

Hence,  when  the  two  given  circles  are  in  contact,  the  two  transverse 
common  tangents  coincide  and  become  the  common  tangent  at  the  point  of 
contact ;  and  when  the  two  given  circles  intersect,  the  transverse  common 
tangents  cannot  be  drawn. 

5.  If  in  any  two  given  circles  which  are  in  contact  there  be  drawn  two 
parallel  diameters,  the  point  of  contact  and  an  extremity  of  each  diameter 
lie  in  the  same  straight  line. 

Suggestion.  Draw  the  line  of  centres,  and  join  an  extremity  of  each 
diameter  to  the  point  of  contact.  Show  that  the  two  lines  so  drawn  make 
equal  angles  with  the  line  of  centres. 

6.  If  two  circles  have  a  common  point  not  on  the  line  of  centres,  the 
circles  must  intersect  at  that  point,  and  also  at  another  point  which  is 
situated  symmetrically  with  it  relative  to  the  line  of  centres. 

Suggestion.  The  foot  of  the  perpendicular  from  the  common  point 
to  the  line  of  centres  lies  inside  of  both  circles ;  hence  the  circles  must 
intersect. 

7.  Two  circles  touch  each  other  internally  or  externally  at  the  point  A, 
and  through  A  two  straight  lines  are  drawn  cutting  one  circle  in  P  and  i?, 
respectively,  and  the  other  circle  in  A  and  8.  Show  that  PE  is  parallel 
to  ^.S'. 

8.  Two  circles  touch  one  another  at  the  point  A^  and  have  a  common 
tangent  meeting  them  at  the  points  B  and  C,  respectively.  Show  that  the 
circle  whose  diameter  is  JBO  passes  through  A.  Show  also  that  if  the 
lines  BA  and  CA  are  produced  to  cut  the  circles  again  at  C  and  5', 
respectively,  the  lines  BB'  and  CC  will  be  diameters. 

Suggestion.  If  the  common  tangent  at  A  intersects  the  common 
tangent  BC  at  M,  MA  =  MB  =  MG  (Art.  194).  Therefore  Z  5^(7  is  a 
right  angle. 


134  ELEMENTARY  GEOMETBT  [Chap.  II 


MISCELLANEOUS    EXERCISES 

1.  All  chords  of  the  greater  of  two  given  concentric  circles  which  are 
tangent  to  the  smaller  are  equal. 

2.  What  is  the  locus  of  the  centres  of  circles  touching  two  given 
straight  lines  which  intersect  ? 

3.  Describe  a  circle  of  given  radius  which  shall  touch  two  given  inter- 
secting straight  lines.     Show  that  there  are  four  such  circles. 

4.  A  circle  is  described  on  the  radius  of  another  as  diameter.  Prove 
that  any  chord  of  the  greater  circle  drawn  through  their  point  of  contact 
is  bisected  by  the  lesser  circle. 

6.  Through  a  point  of  intersection  of  two  given  circles  draw  the 
greatest  possible  line-segment  which  is  terminated  both  ways  by  the 
circles. 

6.  If  AB  and  CD  are  two  equal  chords  in  a  circle,  prove  that  of  the 
two  pairs  of  straight  lines  AD,  BC,  and  AC,  BD,  one  pair  are  equal  and 
the  other  parallel. 

7.  Given  two  circles  and  a  tangent  to  each,  these  being  parallel ;  if 
the  points  of  contact  of  the  tangents  be  joined  by  a  straight  line,  the  tan- 
gents at  the  points  where  this  straight  line  cuts  the  circles  a  second  time 
are  also  parallel. 

8.  Two  radii  of  a  circle  are  at  right  angles  and  when  produced  are  cut 
by  a  straight  line  which  touches  the  circle.  Show  that  the  other  tangents 
drawn  from  the  points  of  intersection  with  the  radii  are  parallel. 

9.  If  two  circles  touch  each  other  and  a  chord  be  drawn  through  the 
point  of  contact,  the  tangents  at  the  other  points  where  the  chord  meets 
the  circles  are  parallel. 

10.  From  all  the  points  of  a  circle  equal  and  parallel  line-segments  are 
drawn  in  the  same  direction.     What  is  the  locus  of  their  extremities  ? 

11.  From  any  point  within  a  circle  straight  lines  are  drawn  to  the 
circle  ;  show  that  the  locus  of  their  mid-points  is  a  circle.  What  would 
be  the  locus  of  the  mid-points  if  the  lines  were  drawn  from  a  point  on  or 
outside  the  circle  ? 

12.  Let  AB  be  any  diameter  of  a  circle  whose  centre  is  0,  and  let  C 
be  any  point  on  this  diameter  produced.  Through  C  draw  any  secant 
cutting  the  circle  at  D  and  E.  If  the  exterior  part  CD  of  this  secant  is 
equal  to  a  radius  of  the  circle,  show  that  the  angle  EOA  is  three  times 
the  angle  DOB. 


209j  THE  CIRCLE  135 

13.  What  is  the  locus  of  the  mid-points  of  equal  chords  of  a  circle  ? 

14.  If  two  equal  chords  of  a  circle  are  produced  to  meet  outside  the 
circle,  prove  that  the  exterior  parts  are  equal.  What  is  the  correspond- 
ing property  when  the  chords  intersect  inside  the  circle  ?  Notice  the 
principle  of  continuity. 

15.  What  is  the  locus  of  the  centres  of  circles  of  constant  radius  which 
touch  a  given  circle  ? 

16.  A  straight  line  is  drawn  intersecting  two  concentric  circles.  Show 
that  the  line-segments  intercepted  between  the  circles  are  equal. 

17.  Circles  are  described  on  the  sides  of  a  quadrilateral  as  diameters. 
Show  that  the  common  chord  of  the  circles  described  on  two  adjacent 
sides  is  parallel  to  the  common  chord  of  the  other  two  circles. 

18.  If  AB  and  A' B'  are  two  equal  line-segments  lying  in  a  plane,  but 
not  parallel,  find  a  point  0  such  that  if  the  line  J.0  be  rotated  about  it 
through  a  certain  angle,  A  will  coincide  with  A'  and  at  the  same  time 
B  with  B'. 

19.  Three  circles  touch  one  another  externally  (each  touching  the 
other  two),  at  the  points^,  B,  C ;  the  straight  lines  AB,  AC,  are  pro- 
duced to  meet  the  circle  BC  aX  D  and  E.  Show  that  DE  is  a  diameter 
of  this  circle  parallel  to  the  line  of  centres  of  the  other  two  circles. 

20.  If  AB  is  a  fixed  diameter,  and  DE  an  arc  of  constant  length  in  a 
given  circle,  and  the  lines  AE,  BD  intersect  at  P,  show  that  the  angle 
APB  is  constant. 

21.  Three  concurrent  straight  lines  make  fixed  angles  with  each  other. 
If  they  be  moved  so  that  two  of  them  constantly  pass  through  fixed 
points,  the  third  must  also  pass  through  a  fixed  point. 

22.  A  triangle  is  inscribed  in  a  circle.  Show  that  the  sura  of  the 
angles  contained  in  the  three  arcs  subtended  by  the  sides  is  equal  to  four 
right  angles. 

23.  If  ABC  is  any  triangle  and  a  circle  is  described  through  the  ver- 
tices B  and  O,  cutting  the  sides  BA  and  CA,  at  the  points  P  and  Q, 
prove  that  PQ  is  parallel  to  a  fixed  straight  line. 

24.  If  ^  is  a  point  on  one  of  the  diagonals  ^C  of  a  parallelogram 
ABCD,  and  circles  are  described  about  DEA  and  BEC,  show  that  the 
other  point  of  intersection  of  these  two  circles  must  lie  on  BD. 

25.  If  through  P,  any  point  on  one  of  two  circles  which  intersect  at  A 
and  P,  the  straight  lines  PA  and  PB  are  drawn  to  meet  the  other  circle 
at  Q  and  P,  prove  that  the  arc  QB  is  of  constant  length,  or  that  the 
length  of  QB  is  independent  of  the  position  of  the  point  P. 


136  ELEMENTARY  GEOMETRY  [Chap.  II 

SUMMARY  OF  CHAPTER  II 

1.  Definitions. 

(1)  Circle,  centre,  radius,  diameter.    §  147.     See  also  p.  20. 

(2)  Arc  of  a  Circle  —  any  portion  of  a  circle  terminated  by  two 

points.    §  153. 

(3)  Chord  of  a  Circle  —  the  straight  line  joining  two  points  of  a 

circle.     §  153. 

(4)  Conjugate  Arcs  of  a  Circle  —  two  arcs  which  together  make  up 

the  whole  circle.     §  153. 

(5)  Semicircle  —  the  arc  of  a  circle  subtended  by  a  diameter.     §  154. 

(6)  Segment  of  a  Circle  —  the  figure  formed  by  an  arc  and  its  sub- 

tending chord.     §  156. 

(7)  Sector  of  a  Circle  —  the  figure  formed  by  an  arc  and  the  two 

radii  to  its  extremities.     §  156. 

(8)  Inscribed  Figure  —  a  rectilinear  figure  is  said  to  be  inscribed  in  a 

circle  when  its  vertices  lie  on  the  circle.     §  177. 

(9)  Circumscribed  Figure  —  a  rectilinear  figure  is  said  to  be  circum- 

scribed about  a  circle  when  its  sides  are  all  tangent  to  the  circle. 
§196. 

(10)  Concyclic  Points  —  points  which  lie  on  the  same  circle.     §  184. 

(11)  Secant  of  a  Circle  —  an  unlimited  straight  line  which  intersects  a 

circle  in  two  points.     §  185. 

(12)  Tangent  to  a  Circle  —  a  straight  line  which  meets  a  circle,  but 

which  when  produced  does  not  cut  it.     §  186. 

(13)  Chord  of  Contact  — i\i&  straight  line  joining  the  points  of  contact 

of  two  tangents.     §  194. 

(14)  Circles  in  Contact  —  two  circles  which  have  a  common  tangent  at 

a  common  point.     §  205. 

(15)  Direct  and  Transverse  Common  Tangents.     See  Ex.  2,  p.  133. 

(16)  Principle  of  Continuity  —  the  principle  which  asserts  that  a  rela- 

tion among  the  parts  of  a  geometrical  figure,  once  true  and 
properly  interpreted,  remains  true  when  the  figure  changes 
continuously  from  one  form  to  another,  subject  to  the  con- 
ditions under  which  it  was  first  described.     §  204. 

2.  Problems. 

(1)  To  find  the  centre  of  a  circle  which  passes  through  three  given 

points,  or  to  circumscribe  a  circle  about  a  given  triangle.    §  151. 

(2)  To  inscribe  a  circle  in  a  given  triangle.     §  196. 

(3)  To  draw  a  tangent  to  a  circle  from  a  given  point  on  it.     §  187. 


Summary]  THE  CIRCLE  137 

(4)  T6  draw  a  tangent  to  a  circle  from  a  given  point  without  it. 

§  193. 

(5)  On  a  given  line-segment  to  construct  an  arc  of  a  circle  which  shall 

contain  a  given  angle.     §  199. 

(6)  To  draw  a  common  tangent  to  two  given  circles,  each  of  which 

lies  wholly  outside  of  the  other.     §  209. 

3.  Theorems  on  the  Coincidence  of  Circles. 

(1)  Two  circles  in  a  plane  which  have  the  same  centre  and  equal 

radii  coincide  throughout.    §  148. 

(2)  Two  circles  in  a  plane  which  have  the  same  centre  and  one  point 

in  common  coincide  throughout.     §  148. 

(3)  Two  circles  which  have  equal  radii  can  be  made  to  coincide,  and 

hence  are  identically  equal ;  and,  conversely,  equal  circles  have 
equal  radii.     §  148. 

(4)  Through  three  points  not  in  the  same  straight  line,  one  and  only 

one  circle  can  be  described.     §  151. 

(5)  Two  circles  which  coincide  at  three  points  coincide  throughout. 

§  151. 

(6)  Two  different  circles  can  have  at  most  two  points  in  common. 

§151. 

(7)  A  straight  line  can  intersect  a  circle  in  at  most  two  points.    §  152. 

4.  Theorems  on  Arcs,  their  Chords  and  Angles. 

(1)  In  equal  circles  or  in  the  same  circle : 

(a)  Equal  angles  at  the  centre  are  subtended  by  equal  arcs ; 

and  of  two  unequal  angles  the  greater  is  subtended  by 

the  greater  arc.     §§  157,  158. 
(6)  Equal  arcs  subtend  equal  angles  at  the  centre  ;  and  of  two 

unequal  arcs  the  greater  subtends  the  greater  angle  at 

the  centre.     §§  159,  160. 
(c)    Equal  arcs  are  subtended  by  equal  chords  ;  and  of  two 

unequal  minor  arcs,  the  greater  is  subtended  by  the 

greater  chord.      §§  161,  162. 
(df)  Equal  chords  subtend  equal  arcs ;    and  of  two  unequal 

chords,  the  greater  subtends  the  greater  arc.      §  163. 
(e)    Equal  chords  subtend  equal  angles  at  the  centre,  and  of 

two  unequal  chords,  the  greater  subtends  the  greater 

angle.     §  164. 

(2)  An  angle  whose  vertex  is  on  a  circle  equals  half  the  angle  at  the 

centre  subtended  by  the  same  arc.     §  178. 

(3)  The  angle  in  a  semicircle  is  a  right  angle.     §  179. 


188  ELEMENTARY  GEOMETRY  [Chap.  II 

(4)  An  angle  in  an  arc  greater  than  a  semicircle  is  acute,  and  an 

angle  in  an  arc  less  than  a  semicircle  is  obtuse.     §  180. 

(5)  All  angles  in  a  circle  subtended  by  the  same  arc,  or  by  equal 

arcs,  are  equal.     §  181. 

(6)  Equal  angles  on  the  same  base,  and  on  the  same  side  of  it,  have 

their  vertices  on  an  arc  of  a  circle  of  which  the  given  base  is  the 
chord.     §  182. 

5.  Theorems  on  Chords  in  relation  to  the  Centre. 

(1)  The  perpendicular  bisector  of  any  chord  of  a  circle  passes  through 

the  centre.     §  155. 

(2)  The  line  drawn  from  the  centre  perpendicular  to  a  chord  bisects 

the  chord,  and,  if  produced,  bisects  the  arc  subtended  by  the 
chord.     §  165. 

(3)  The  straight  line  drawn  from  the  centre  to  the  mid-point  of  a 

chord  is  perpendicular  to  the  chord.     §  166. 

(4)  The  mid-points  of  a  system  of  parallel  chords  all  lie  on  a  diameter 

perpendicular  to  the  chords.     §  167. 

(5)  The  diameter  which  bisects  a  chord  also  bisects  the  angle  at  the 

centre  subtended  by  the  chord.     §  169. 

(6)  In  the  same  circle  or  in  equal  circles  : 

(a)  Equal  chords  are  equidistant  from  the  centre  ;  and  of 
two  unequal  chords,  the  greater  is  nearer  to  the  centre 
than  the  less.     §  170.     And,  conversely  — 

(6)  Chords  equidistant  from  the  centre  are  equal ;  and  of 
two  chords  unequally  distant,  the  one  nearer  the  centre 
is  the  greater.     §  171. 

(7)  A  diameter  is  the  greatest  chord  that  can  be  drawn  in  any  circle. 

§172. 

6.  Theorems  relating  to  Tangents. 

(1)  The  perpendicular  to  a  diameter  of  a  circle  at  one  extremity  is  a 

tangent  to  the  circle  ;  and  any  other  straight  line  through  that 
extremity  will  cut  this  circle  at  a  second  point.     §  187. 

(2)  Any  tangent  to  a  circle  is  perpendicular  to  the  radius  drawn  to 

the  point  of  contact.     §  189. 

(3)  The  centre  of  a  circle  lies  on  the  perpendicular  to  a  tangent  drawn 

from  the  point  of  contact.     §  190. 

(4)  The  straight  line  drawn  from  the  centre  of  a  circle  perpendicular 

to  a  tangent  meets  it  at  the  point  of  contact.     §  191. 

(5)  The  tangent  to  a  circle  at  the  mid-point  of  any  arc  is  parallel  to 

the  chord  of  the  arc.    §  192. 


Summary]  THE  CIRCLE  139 

(6)  At  any  point  of  a  circle  there  can  be  drawn  one  and  only  one 

tangent.     §  188. 

(7)  From  a  point  outside  of  a  circle  there  can  be  drawn  two  and  only 

two  tangents.     §  195. 

(8)  The  two  tangents  to  a  circle  drawn  from  any  outside  point  are 

equal,  and  the  straight  line  from  the  point  to  the  centre  of  the 
circle  bisects  the  angle  between  them.     §  194. 

(9)  The  angle  formed  by  a  tangent  to  a  circle  and  a  chord  through  its 

point  of  contact  equals  half  the  angle  at  the  centre,  subtended 
by  the  intercepted  arc  ;  or  is  equal  to  the  angle  in  the  segment 
on  the  opposite  side  of  the  chord.     §§  197,  198. 

7.  Theorems  on  Two  Circles. 

(1)  The  straight  line  joining  the  centres  of  two  intersecting  circles 

bisects  their  common  chord  at  right  angles.     §  168. 

(2)  When  two  circles  touch  each  other,  their  point  of  contact  and 

their  two  centres  are  collinear.     §  206. 

(3)  Two  circles  which  touch  each  other  can  have  no  other  point  in 

common  than  the  point  of  contact.     §  207. 

(4)  When  two  circles  have  a  common  point  on  their  line  of  centres, 

they  must  have  a  common  tangent  at  that  point.    §  208. 

8.  Miscellaneous  Theorems. 

(1)  Of  all  straight  lines  which  can  be  drawn  to  a  circle  from  a  point 

within  it,  not  the  centre,  the  greatest  is  that  which  passes 
through  the  centre,  and  the  least  is  that  which,  if  produced 
backward,  would  pass  through  the  centre  ;  and  of  any  two 
others,  the  greater  is  that  which  makes  the  less  angle  with  the 
diameter  through  the  point.     §  173. 

(2)  From  any  point  within  a  circle  two  equal  straight  lines  can  be 

drawn  to  the  circle  ;  these  make  equal  angles  with  the  diameter 
through  the  point.     §  174. 

(3)  The  circle  described  upon  the  hypotenuse  of  a  right  triangle  as 

diameter  passes  through  the  vertex  of  the  right  angle.     §  183. 

(4)  The  opposite  angles  of  any  convex  quadrilateral  inscribed  in  a 

circle  are  supplementary.     §  184. 
Conversely,  if  two  opposite  angles  of  a  convex  quadrilateral  are 
supplementary,  its  vertices  are  concyclic.     §  184. 

(5)  The  angle  formed  by  two  secants  of  a  circle  is  equal  to  the  sum 

or  the  difference  of  the  angles  whose  vertices  are  on  the  circle 
and  which  are  subtended  by  the  intercepted  arcs,  according  as 
the  secants  intersect  within  or  without  the  circle.     §§  200,  202. 


CHAPTER  III 
SIMILAR  RECTILINEAR  FIGURES 

Section  I 

MEASUREMENT,   RATIO,   AND  PROPORTION 
1.    On  Measurement 

210.  If  you  are  asked  to  measure  a  stick  or  a  piece  of  cloth, 
you  first  select  a  magnitude  which  is  well  known,  say  a  foot- 
rule  or  a  yard-stick,  and  then  compare  the  length  of  the  given 
stick  or  piece  of  cloth  with  the  chosen  unit  length.  You  find 
perhaps  that  the  given  stick  is  four  feet  long,  or  that  the  piece 
of  cloth  is  five  yards  long. 

What  is  meant  by  saying  that  the  stick  is  four  feet  long  ? 
Simply  that  the  length  of  the  stick  is  four  times  the  length  of 
the  foot-rule. 

211.  Definition.  To  measure  anything  is  to  find  out  by 
experiment  how  many  times  it  will  contain  a  chosen  unit. 

If  one  magnitude  contains  another  an  integral  number  of 
times,  the  first  is  called  a  multiple  of  the  second,  and  the 
second  a  measure  of  the  first. 

The  measure  of  any  magnitude  is  a  number  expressing  how 
many  times  it  contains  the  unit  of  measure. 

212.  The  unit  of  measure  and  the  thing  to  be  measured 
must  of  course  be  magnitudes  of  the  same  kind. 

In  the  case  mentioned  we  had  two  magnitudes  of  the  same 

140 


210-214J  SIMILAR  BECTILINEAB   FIGUBES  141 

kind,  viz.  the  length  of  the  stick  and  the  length  of  the  foot- 
rule,  and  we  say  that  the  measure  of  the  former  is  four  when 
the  latter  is  taken  for  unit. 

Exercise.  What  is  the  measure  of  12  feet  when  a  yard  is  taken  for 
unit  ?  Of  a  dollar  when  a  dime  is  taken  for  unit  ?  Of  a  mile  when  a 
foot  is  taken  for  unit  ?    Of  68  when  4  is  taken  for  unit  ? 

213.  If  two  magnitudes  are  multiples  of  the  same  magni- 
tude, or,  in  other  words,  if  two  magnitudes  have  a  common 
measure,  they  are  said  to  be  commensurable. 

The  numbers  21  and  33  have  a  common  measure  3 ;  the  num- 
bers 17  and  43  have  a  common  measure  1 ;  the  numbers  4^ 
and  3^  have  a  common  measure  |  or  Jj. 

Two  magnitudes  which  have  no  common  measure  are  called 
incommensurable. 

As  examples  of  incommensurable  magnitudes  the  following 

may  be  mentioned : 
(1)  The  circumference  and  diameter  of  a  circle ;  (2)  the  side 

and  diagonal  of  a  square. 
The  numbers  7  and  VIl,  1  and  V2  are  incommensurable. 

214.  The  standard  unit  for  measuring  lengths  in  the  English 
or  common  system  is  the  yard,  and  in  the  French  system,  which 
is  coming  to  be  used  almost  universally  for  scientific  purposes, 
the  standard  unit  is  the  metre. 

These  units  are  defined  to  be  lengths  equal  to  the  lengths  of 
certain  pieces  of  metal  which  are  carefully  preserved,  the  first 
in  London,  the  second  in  Paris. 

For  making  small  measurements  these  units  are  too  large 
for  convenient  use,  and  so  fractional  parts  of  them  are  used  in 
such  cases ;  for  example,  '  a  foot '  is  J  of  a  yard ;  '  an  inch '  is 
jL  of  a  foot  or  -j^  of  a  yard ;  ^  a  centimetre '  is  y^-g-  of  a  metre, 
and  so  on. 

On  the  other  hand,  for  making  large  measurements  we  use 
'  a  mile,'  which  equals  1760  yards,  and  so  on. 


142  ELEMENTARY  GEOMETRY  [Chap.  Ill 

215.  For  measuring  angles  the  standard  unit  is  the  right 
angle,  defined  in  Article  15,  page  9. 

This  unit  again  is  too  large  for  convenient  use,  and  a  smaller 
unit  is  commonly  employed  in  making  actual  measurements, 
namely,  a  degree. 

A  degree  is  defined  to  be  -^^  of  a  right  angle. 

A  minute  is  -^^  of  a  degree,  and  a  second,  ^-^  of  a  minute. 


EXERCISES 

1.  How  many  degrees  in  two  right  angles  ?  In  two-thirds  of  a  right 
angle  ?    In  one-fifth  of  a  right  angle  ? 

2.  How  many  degrees  in  the  sum  of  the  angles  of  a  triangle  ? 

3.  How  many  degrees  in  each  of  the  angles  of  an  equilateral  triangle  ? 

4.  How  many  degrees  in  each  of  the  angles  of  an  isosceles  right  triangle  ? 

5.  One  angle  of  a  right  triangle  equals  30°.  How  many  degrees  in  each 
of  the  others  ? 

6.  The  vertical  angle  of  an  isosceles  triangle  equals  18°.  How  many 
degrees  in  each  of  the  base  angles  ? 

7.  One  of  the  base  angles  of  an  isosceles  triangle  equals  50°.  How 
large  is  the  vertical  angle  ? 

8.  One  angle  of  a  parallelogram  equals  45^  What  is  the  measure  of 
each  of  the  other  angles  ? 

2.   On  Eatio 

216.  When  we  speak  of  the  ratio  of  one  quantity  to  another, 
we  have  in  mind  their  relative  magnitude.  By  this  we  mean 
not  how  much  the  one  is  greater  or  less  than  the  other,  but  how 
many  times  the  one  is  as  great  as  the  other. 

A  ratio  can  be  expressed  only  between  two  quantities  of  the 
same  kind. 

When  dealing  with  numbers  we  say  that  the  ratio  of  one  to 
another  is  the  quotient  arising  from  dividing  the  first  by  the 


215-219]  SIMILAR  RECTILINEAR  FIGURES  143 

second,  since  the  quotient  expresses  how  many  times  the  one 
is  as  great  as  the  other. 

Thus  the  ratio  of  8  to  4  is  2,  the  ratio  of  7  to  3  is  2i  the 
ratio  of  5  to  9  is  f ,  etc. ;  but  the  division  of  one  geometrical 
magnitude  by  another  has  no  meaning  unless  the  first  is  an 
exact  multiple  of  the  second,  or  until  a  meaning  is  assigned 
by  definition.  Consequently,  the  numerical  value  of  the  ratio 
of  two  such  magnitudes  must  sometimes  be  got  at  in  a  round- 
about way. 

217.  The  same  two  quantities  A  and  B  have  two  different 
ratios,  viz.  the  magnitude  of  A  as  compared  with  B,  and  the 
magnitude  of  B  as  compared  with  A. 

The  first  is  expressed  A  :  B,  and  should  be  read  '  the  ratio  of 
A  to  ^ ' ;  the  second  is  expressed  B :  A  and  should  be  read 
*  the  ratio  of  B  to  A.' 

In  any  ratio  the  first  term  is  called  the  antecedent,  the  second 
term,  the  consequent. 

218.  The  following  postulates  will  serve  to  add  definiteness 
to  the  meaning  of  the  term  '  ratio.' 

Postulate  6.  If  P  and  Q  are  any  two  equal  magnitudes  and 
R  is  a  third  magnitude  of  the  same  kind,  then  the  ratio  of  P  to  R 
is  equal  to  the  ratio  of  Q  to  R,  i.e.  if  P  =  Q,  then  P:  R=  Q:  R; 
and,  conversely,  if  P  and  Q  are  such  that  P :  R  =  Q :  R,  then 
P=Q. 

Postulate  7.  If  P  and  Q  are  two  unequal  magnitudes,  and 
R  is  a  third  magnitude  of  the  same  kind,  then  the  ratio  of  Pto  R 
is  greater  or  less  than  the  ratio  of  Q  to  R,  according  as  P  is 
greater  or  less  than  Q. 

219.  In  order  to  show  how  the  ratio  of  one  geometrical 
magnitude  to  another  can  be  expressed  numerically,  it  is  neces- 
sary to  consider  two  distinct  cases. 


144  ELEMENTARY  GEOMETRY  [Chap. Ill 

(a)  When  the  two  given  magnitudes  have  a  common 
measure,  or  are  commensurable. 

(6)  When  the  two  given  magnitudes  have  no  common 
measure,  or  are  incommensurable. 


(a)     RATIO    OF    COMMENSURABLE    MAGNITUDES 

220.  Definition.  The  ratio  of  two  commensurable  magni- 
tudes is  the  ratio  of  their  numerical  measures  by  a  common 
unit. 

Let  A  and  B  be  any  two  magnitudes  of  the  same  kind  which 
have  a  common  measure  S,  and  let  S  be  contained  in  ^  m 
times,  and  in  5  n  times,  so  that  m  is  the  measure  of  A,  and  n 
the  measure  of  B,  by  the  common  unit  S. 

Then  by  definition 

A  :  B  =  m  :  n. 

But  the  ratio   of   the  number  m  to  the  number  n  is  the 

quotient  — 
n 

Therefore  A:B='^' 

n 

221.  If  instead  of  A  and  B  we  take  any  equimultiples  of 
these  magnitudes,  say  pA  and  pB,  their  measures  by  the 
common  unit  S  would  be  pm  and  pn. 

Then  by  definition 

pA  :  pB  —  pm :  pn 
pm     m 
~ pn  ~~  n' 

But  —  is  the  ratio  of  A  to  B. 
n 

Therefore,  the  ratio  of  any  two  commensurable  magnitudes  is 

equal  to  the  ratio  of  any  equimultiples  of  those  magnitudes,  taken 

in  the  same  order. 

The  ratio  of  5  ft.  to  7  ft.  is  the  same  as  the  ratio  of  10  ft.  to  14 
ft.,  or  of  25  ft,  to  35  ft. 


219-224]  SIMILAR  BECTILINEAB  FIGURES  145 

222.  Again,  if  instead  of  the  common  measure  S  we  use  a 

different  common  measure,  say  — ,  so  that  the  measure  of  A 

t 

becomes  tm  and  the  measure  of  B  becomes  tn,  then 

A:  B  =  tm:tn 
_tm  __m 
tn      n 

That  is,  the  ratio  of  A  to  B  is  not  altered  by  a  change  in  the 
unit  of  measure. 

The  ratio  of  5  ft.  to  7  ft.  is  not  altered  if  those  lengths  are 
expressed  in  inches,  or  in  yards,  instead  of  feet. 

223.  For  convenience    we    frequently   write  A:  B   in   the 

form  —  even  when  A  and  B  represent  geometrical  magnitudes ; 
B 

but  the  symbol  so  used  should  always  be  read  as  'the  ratio 
of  A  to  B/  and  should  not  be  confounded  with  an  ordinary 
fraction,  or  symbol  of  division. 

/    i  Tip 

Thus,      ^     ^  expresses  the  ratio  of  Z  ABC  to  Z  PQR,  and 
Z  PQR 

may  be  written  in  that  form  or  in  the  usual  ratio  form 

Z  ABC :  Z  PQR. 

If  Z  AOB  is  at  the  centre  of  a  circle  subtended  by  the  arc 

AB,  and  Z.ACB  is  at  a  point  of  the  circle  subtended  by  the 

.,        ZAOB     2        -,   ZACB     1   ,,  .      .^„ 

same  arc,  then  — — — -— =  -  and  — — — -  =  -,  the  angle  AUB 
Z  ACB     1  ZAOB     2 

being  taken  for  unit  of  measure  in  each  case.  (Art.  178.) 

224.  Whenever  the  first  of  the  two  given  magnitudes  is  an 
exact  multiple  of  the  second,  the  second  may  be  taken  as  the 
common  unit  of  measure ;  and  then  the  ratio  of  the  first  to  the 
second  is  equal  to  the  measure  of  the  first  by  the  second. 

The  ratio  of  8.  to  4  is  2,  of  20  to  5  is  4,  of  the  diameter  to  a 
radius  of  a  circle  is  2,  of  the  perimeter  to  a  side  of  an  equi- 
lateral triangle  is  3,  etc. 

L 


146  ELEMENTARY  GEOMETRY  [Chap.  Ill 

(b)    RATIO    OF    INCOMMENSURABLE    MAGNITUDES 

225.  Next  suppose  that  A  and  B  are  two  magnitudes  such 
that  no  unit  however  small  will  measure  them  both  integrally, 
that  is,  suppose  that  A  and  B  are  incommensurable. 

In  that  case  the  ratio  of  ^  to  ^  cannot  be  expressed  either 
as  a  whole  number  (Art.  224)  or  as  a  fraction  whose  numerator 
and  denominator  are  both  whole  numbers  (Art.  220),  since  they 
have  no  common  measure. 

But  while  the  ratio  of  two  such  magnitudes  does  not  abso- 
lutely equal  any  integer  or  common  fraction,  it  is  always  pos- 
sible to  find  a  common  numerical  fraction  which  will  differ  in 
value  from  their  ratio  by  less  than  any  assigned  quantity  how- 
ever small,  as  we  shall  now  proceed  to  show. 

226.  It  is  necessary  before  going  further  to  introduce  a  new 
idea,  which  we  shall  do  by  means  of  an  illustration. 

A  boy  is  to  walk  from  P  to  Q,  a  distance  of  two  miles.  He 
goes  half  the  distance  the  first  hour,  half  the  remaining  dis- 
tance the  second  hour, 

half  the  remaining  dis-    P>— ■ ' ^ ' — ' — Q 

tance  the   third    hour, 

and  so  on.  Would  he  ever  absolutely  reach  his  distination  ? 
Could  you  fix  a  point  between  P  and  Q,  as  near  as  you  like 
to  Q,  beyond  which  he  would  not  pass  in  time  ?  Would  he 
ever  get  beyond  Q  ?  . 

In  the  first  hour  the  boy  would  go  1  mile,  in  the  second,  \ 
mile ;  in  the  third,  \  mile ;  in  the  fourth,  i  mile ;  and  so  on. 

Take  these  numbers  1,  \,  J,  \,  yV?  A?  '"^  ^^^  ^^*  '^  represent 
the  sum  of  n  of  them. 

Then  when  w  =  1,  iS  =  1 ; 

when  n  =  2,  /S'  =  l  +  i  =  f; 
when  71  =  3,  >S  =  l-h|  +  i  =  }; 

when  n  =  4,  ^  =  l  +  i-|-i  +  i  =  V; 
and  so  on. 


226-228]  SIMILAR  RECTILINEAR  FIGURES  147 

Here  /S  is  a  variable  quantity  dependent  upon  the  value  of 
the  integer  n.  As  n  increases,  S  increases.  By  taking  n  large 
enough,  we  can  make  S  as  nearly  equal  to  the  fixed  number 
two  as  we  please,  but  it  can  never  be  absolutely  equal  to  two, 
and  can  never  exceed  two. 

The  fixed  quantity  two  is  called  the  limit  of  the  variable 
quantity  S. 

227.  Definition.  If  a  quantity  is  made  to  vary,  by  chang- 
ing in  some  definite  way  another  quantity  on  which  it  depends, 
and  its  value  approaches  a  fixed  •  quantity  nearer  than  for  any 
assignable  difference,  though  it  cannot  be  made  absolutely  iden- 
tical with  it,  this  fixed  quantity  is  called  the  limit  of  the  vari- 
able quantity. 

In  the  illustration  of  Article  226,  the  variable  quantity  S  is 
called  the  dependent  variable,  and  the  quantity  n  on  which  it 
depends  the  independent  variable. 

228.  Now  let  A  and  B  be  two  incommensurable  magnitudes, 
for  example,  two  line-segments.  Then  AiBissi  so-called  incom- 
mensurable ratio.     Take  a  unit       .  ,      . 

length   X,  which  will   measure  ** 

B  integrally  n  times,  and  apply  - 

it  as  often  as  possible  to  A,  say  * 

m  times,  m  and  n  being  thus  both  integers.     Since  A  and  B 

are  incommensurable,  the  unit  x  will  not  measure  A  exactly, 

and  there  must  be  a  remainder  which  is  less  than  x. 

If  we  repeat  the  process,  taking  for  unit  a  measure  of  B 
smaller  than  x,  then  m  and  n  will  ordinarily  be  greater  than 
before,  and  the  remainder  for  A  will  again  be  smaller  than  the 
chosen  unit.  By  taking  the  unit  small  enough,  we  can  make 
this  remainder  as  small  as  we  please,  though  it  can  never  be 
made  to  disappear  altogether. 

When  the  chosen  unit  of  measure  is  x,  A  is  greater  than  mx, 
but  less  than  (m  +  1)  x,  while  B  equals  nx. 


148  ELEMENTARY  GEOMETRY  [Chap.  Ill 

Therefore   A  :  B  is   greater    than   mx  :  nx,   but   less   than 
(m  +  l)x:  7ix,  (Art.  218,  Post.  7.) 

or  A  :  B  is  greater  than  m :  n,  but  less  than  (m  + 1) :  n. 

That  is,       A:  B  lies  between  —  and  ^^^^±1. 

n  n 

Now  -  differs  from  ^^i+i  by  -• 
n  n  n 

Therefore   A  :  B  differs  from  —  by  less  than  — 

n  n 

By  taking  the  unit  of  measure  x  small  enough,  n  can  be 

made  as  great  as  we  please,  i.e.  -  as  small  as  we  please,  and, 

n 

consequently,  smaller  than  any  number  you  may  choose  to 
name. 

Therefore  by  choosing  x  small  enough,  we  can  always  find  a 

fractional  number   —  (whose  numerator  and  denominator  are 
n 

both  integers)  which  differs  in  value  from  the  incommensura- 
ble ratio  ^ :  jB  by  less  than  any  number  you  choose  to  name, 
no  matter  how  small,  i.e.  by  less  than  any  assigned  quantity. 

The  variable  fraction  —  approaches  a  limit,  and  this  limit 
n 

is  taken  as  the  numerical  value  of  the  ratio  of  A  to  B. 


229.  Definition.  The  ratio  of  two  incommensurable  mag- 
nitudes A  and  B  is  the  limit  which  the  fraction  —  (defined 

n 

in  Article  228)  approaches  as  the  unit  x  upon  which  it  depends 
is  indefinitely  dirainished. 

If  A  and  B  are  commensurable  as  in  Article  220,  some  one 
of  the  fractions  found  in  proceeding  to  the  limit  will  itself 
be  their  ratio. 

230.  General  Theorem  on  Limits.  If  there  are  two  varia- 
ble quantities  dependent  cm  the  same  quantity  in  such  a  way  that 


228-232]  SIMILAR  RECTILINEAR  FIGURES  149 

they  remain  always  equal  while  each  approaches  a  limits  then 
their  limits  are  equal. 

For,  if  the  limits  are  not  equal,  they  must  differ  by  some 
finite  quantity,  say  k;  and  since  each  variable  approaches 
indefinitely  near  to  its  own  limit,  the  limits  differing  by  k, 
the  two  variables  must  come  finally  to  differ  by  a  finite  quan- 
tity, which  is  contrary  to  the  assumption  that  they  remain 
always  equal.  Therefore  the  limits  cannot  differ  by  any 
finite  quantity ;  in  other  words,  the  limits  are  equal. 

231.  By  the  definition  given  in  Art.  220,  the  ratio  of  two 
commensurable  magnitudes  can  always  be  expressed  as  the 
ratio  between  two  fixed  integral  numbers  (viz.  their  measures 
by  a  common  unit) ;  and  by  the  definition  of  Article  229,  the 
ratio  of  two  incommensurable  magnitudes  is  the  limit  of  the 
ratio  of  two  variable  integral  numbers. 

Hence  the  properties  of  the  ratios  of  integral  numbers  can 
be  applied  directly  to  geometrical  magnitudes. 

3.    Proportion 

232.  Definition.  If  four  quantities  are  such  that  the  ratio 
of  the  first  to  the  second  equals  the  ratio  of  the  third  to  the 
fourth,  these  four  quantities  are  said  to  form  a  proportion. 

That  is,  if  the  four  quantities,  a,  b,  c,  d,  are  such  that  a:  b  = 
c :  d,  then  they  form  a  proportion,  or  are  in  proportion ;  or,  the 
terms  a  and  b  are  proportional  to  c  and  d. 

A  proportion  may  be  written  in  the  form  a :  b  =  c :  d,  ot  in 

the  fractional  form  -  =  - ,  and  should  be  read  '  the  ratio  of  a 
b      d' 

to  b  equals  the  ratio  of  c  to  d,'  or  '  a  is  to  b  as  c  is  to  d.' 

Sometimes,  in  expressing  the  equality  of  two  ratios,  the 
double  colon  (: :)  is  used  instead  of  the  ordinary  sign  of  equal- 
ity.    Thus  a:b::c:d  means  the  same  as  a  :  6  =  c  :  d. 

In  a  proportion  the  first  and  fourth  terms  are  called  the 
extremes,  the  second  and  third,  the  means. 


150  ELEMENTARY  GEOMETRY  [Chap.  Ill 

233.  Theorem.  If  four  numbers  are  in  proportion,  the  prod- 
uct of  the  extremes  equals  the  product  of  the  means. 

For  ii  a:b  =c:d,  then  -  =  -,  and  multiplying  both  sides  of 
this  equation  by  bd,  we  obtain  ad  =  be. 

Conversely.  If  the  product  of  two  numbers  equals  the  prod- 
uct of  two  others,  one  pair  can  be  the  extremes  and  the  other  pair 
the  means  in  a  proportion. 

If  ad  =  be,  dividing  by  bd  gives  -  =  - or  a:6  =  c:d 

b     d 

234.  The  property  expressed  in  the  above  theorem,  namely, 
that  the  product  of  the  extremes  equals  the  product  of  the 
means,  may  be  called  the  essential  property  of  a  proportion, 
and  the  terms  of  the  proportion  may  be  written  in  any  order 
which  will  preserve  this  property. 

Thus,  if  (1)  a'.b  =  c:d, 

then  ad  =  be,  and  the  proportion  may  be  written  either  in  the 

form  (2),         b  :a  =  d:  c, 

or  (3),         a:c  =  b:  d, 

since  these  forms  (2)  and  (3)  have  the  same  essential  property 
as  form  (1). 

Form  (2)  may  be  deduced  from  the  essential  property  by 
dividing  both  sides  by  ac  after  interchanging  the  sides. 

Thus  be  =  ad, 

be  _ad 
ac      ac 

Therefore  -  =  -,  ot  b  :  a  =  d  :  c. 

a      c 

And  form  (3)  may  be  deduced  by  dividing  the  essential  prop- 
erty by  cd. 


233-235]  SIMILAR  RECTILINEAR  FIGURES  151 

Thus  ad  =  bCy 

ad  _bc 
cd     cd 

Therefore  -=-,  or  a:c  =  5:d 

c     d 

If  the  two  sides  of  the  proportions  (1),  (2),  and  (3)  are  inter- 
changed, we  have 

(4)  c :  d  =  a  :  b, 

(5)  d:c  =  b:  a, 

(6)  b:d  =  a:c. 

Form  (2)  is  said  to  be  obtained  from  form  (1)  by  inversion, 
and  form  (3)  from  form  (1)  by  alternation. 

EXERCISES 

1.  How  are  (5)  and  (6)  obtained  from  (4)  ? 

2.  Deduce  forms  (4),  (5),  and  (6)  from  the  essential  property. 

235.  It  should  be  remembered  that  the  two  terms  of  any 
ratio  must  be  magnitudes  of  the  same  kind,  since  magnitudes 
of  different  kinds  cannot  be  compared.  But  it  is  not  neces- 
sary that  the  two  ratios  involved  in  a  proportion  should  relate 
to  magnitudes  of  the  same  kind,  since  the  value  of  any  ratio 
is  a  pure  number,  and  the  proportion  merely  expresses  the 
equality  of  the  two  numbers. 

For  example,  a  proportion  among  four  abstract  numbers, 

4  :  6  =  10  :  15, 

may  be  written  in  any  one  of  the  six  forms  of  the  preceding 

article.     But  if  the  terms  involved  are  concrete,  while  we  may 

write 

4  men  :  6  men  =  10  apples  :  15  apx^les, 

or  6  men  :  4  men  =  15  apples  :  10  apples, 

there  would  be  no  meaning  in  the  statement, 

4  men  :  10  apples  =  6  men  :  15  apples, 

since  no  ratio  can  exist  between  magnitudes  of  different  kinds. 


152  ELEMENTARY   GEOMETRY  [Chap.  Ill 

Hence,  when  a  proportion  exists  among  four  quantities,  two 
of  one  sort  and  two  of  another,  care  should  always  be  taken  to 
express  the  ratios  between  the  pairs  of  quantities  of  the  same 
sort. 

236.  Definition.  If  three  quantities,  a,  b,  c,  are  so  related 
that  a:h  =:b  :  c,  they  are  said  to  be  in  continued  proportion,  and 
b  is  called  the  mean  proportional  between  a  and  c,  while  c  is 
called  the  third  proportional  to  a  and  b. 

In  such  a  relation  b'^  =  ac. 

So  also  if  a,  b,  c,  d,  e,  .  .  .  are  so  related  that  a:b  =  b:  c  = 
c :  d,  etc.,  these  quantities  are  said  to  be  in  continued  proportion. 

237.  Theorem.  If  any  four  quantities,  a,  b,  c,  d  are  in  pro- 
portion, then  (1)  a-\-b  :b  =  c-\-d:  d;  (2)  a  —  b:b=c  —  d:d; 
and  (3)  a  +  b  :  a  —b  =  c  -{-  d  :  c  —  d. 

Proof.     (1)  li  a:b  =  c:d,  then 

a_c^ 
b~"d 

Adding  one  to  both  sides  gives 


or 


5  +  1  =  ^  +  1, 
b  d 

a-^b      c-\-d 


(^) 


b  d 

Therefore  a -^b  :b  =  c -\- d:  d. 

This  relation  is  said  to  be  derived  from  the  given  relation 
by  Composition. 

(2)  Similarly,  ^_1  =  £_1, 


or 


b  d 

a  —  b      c  —  d 


b  d 

Therefore  a  —  b  :  b  =  c  —  d  :  d. 


(-B) 


235-238]  SIMILAB  RECTILINEAR  FIGURES  153 

This  relation  is  said  to  be  derived  from  the  given  relation 
by  Division. 

(3)  Dividing  each  member  of  equation  (A)  by  the  corre- 
sponding member  of  equation  (B),  we  have 

a-j-  b      c  +  d 
b     _     d 


b  d 

a  -\-b      c-\-  d 
a—  b      c  —  d 

Therefore       a-{-b:a  —  b  =  c-{-d:c  —  d. 

This  relation  is  said  to  be  derived  from  the  given  relation 
by  Composition  and  Division. 

EXERCISES 

If  a  :  &  =  c  :  d,  show  that 

(1)  ma  :  7ib  =  mc  :  nd, 

(2)  a  :a  ±b  =  c:c  ±d, 

(3)  ma  :  ma  ±nb  =  mc  :mc  ±  nd, 

(4)  a2  .  52  3=  c2  :  (?2, 

(5)  a"  :  ft'*  =  c"  :  d". 

238.  Theorem.  If  three  terms  of  one  proportion  are  equal, 
respectively,  to  the  three  corresjjonding  terms  of  another,  their 
fourth  terms  must  be  equal. 

That  is,  if  a:b  =  c:  d, 

and  a  :  b  =  c' :  d, 

then  c  =  c'. 

Proof.  Since  the  ratios  c  :  d  and  c' :  d  are  each  equal  to  a  :  6, 
therefore,  c  :  d  =  c' :  d, 

and  c  must  equal  c'.  (Art.  218,  Post.  6.) 


154  ELEMENTARY  GEOMETRY  [Chap. Ill 

239.  Theorem.  If  P  :  Q  equals  h :  k,  and  Q :  B  equals 
m  :  n,  then  P :  R  equals  hm  :  kn. 

For,  if|=|thenP  =  |.Q; 

and,  if  ^  =  «  thenQ  =  ™.i?. 

B      n  n 

Therefore,  P=-'-'E,  or  -  =  ^-^,  that  P:R  =  hm'.  kn. 
k    n  R      k '  n 

240.  Theorem.  Tf  aiiy  number  of  ratios  are  equal,  then  the 
sum  of  all  the  antecederits  is  to  the  smn  of  all  the  consequents  as 
any  one   antecedent  is  to  its  consequent. 

Let  tti :  6i  =  ttg :  62  =  cts  :  &3  =  •  ••. 
It  is  required  to  prove  that 

«!  +  0^2  4-  «3  H :  ?>i  +  ?>2  +  ^3  H =  «i '  K  or  <^2 '  h--'- 

Suppose  r  to  be  the  common  value  of  the  given  ratios,  so 

that 

ttj :  bi  =  r,  then  a^  =  rbi, 

ttg :  62  =  ^j  then  a2  =  rb2, 
ttg :  63  =  r,  then  a^  =  rb^, 
etc.,                    etc. 
Adding,  we  have  tti  +  ag  +  ag  +  •••  =  r6i  +  r^a  -\-rb^-] 

=  (&l+?>2  +  &3  +•••)»•• 

Therefore,       a,  +  a^ +  %  +  ■.■  ^^  ^o,  ^  a^^  ,, 

6l+&2  +  ^3+-  ^  ^2 

or         «!  +  «2  +  «3  H '■  bi -\- bz -\- b^  ■] =  a^ :  61  =  ag  ^  ^2  =  •••• 

241.  While  it  requires  four  quantities  to  form  a  proportion, 
we  sometimes  speak  of  one  quantity  being  proportional  to 
another.     When  we  say  that  A  is  proportional  to  B,  we  mean 


239-241]  SIMILAR  RECTILINEAR  FIGURES  155 

that  A  and  B  are  connected  in  some  way  so  that  if  A  changes 
to  A' J  B  will  change  to  a  value  B'  such  that 

A:A'  =  B:B', 

or  A:B  =  A':B'. 

The  same  thing  may  be  expressed  by  saying  that  the  ratio 
A:  B  is  constant. 

Suppose  A:  B  =  m,  3l  constant. 

Then  A  =  mB, 

and  any  change  in  A  will  cause  a  proportional  change  in  B,  or, 
what  is  the  same  thing,  we  may  say  A  varies  as  B. 

If,  on  the  other  hand,  the  product  of  A  and  B  is  constant, 

that  is,  AB  =  m,  a  constant, 

then  any  increase  or  decrease  in'^  will  cause  a  proportional 
decrease  or  increase  in  B,  and  we  say  that  A  is  inversely  pro- 
portional to  B,  or  that  A  varies  inversely  as  B. 

EXERCISES 

1.  What  number  bears  the  same  ratio  to  6  as  5  bears  to  10  ?    As 
3  to  9  ?    As  6  to  9  ? 

2.  Find  the  fourth  proportional  to  5,  8,  10  ;  also  the  fourth  propor- 
tional to  10,  5,  8. 

Definition.     If  a,  b,  c,  d  are  in  proportion,  d  is  called  the  fourth 
proportional  to  a,  6,  and  c. 

3.  It  a:b  =  c:d,  show  that  Sa  +  2b:a  =  Zc-{-2d:c. 

4.  li  a  :b  =  2,  and  6  :  c  =  5,  show  that  a  :  c  =  10. 

5.  Find  the  mean  proportional  between  4  and  9,  between  2  and  32, 
between  4  and  36,  between  5  and  125. 

6.  If  A  varies  as  B,  and  when  ^  =  10,  ^  =  4,  what  is  the  value  of  A 
when  jB  =  14  ? 

7.  If  A  varies  inversely  as  B,  and  when  A  =  2,  B  =  9,  what  is  the 
value  of  B  when  A  =  S? 

8.  If  a  :b  =  b:  c,  show  that  a:c  =  a^  :b^. 


156  ELEMENTARY  GEOMETRY  [Chap.  Ill 

Section  II 

SIMILAR  POLYGONS 

Proposition   I 

242.  A  straight  line  parallel  to  one  side  of  a  triangle 
divides  the  other  two  sides  in  the  same  ratio. 


Let  a  straight  line  parallel  to  the  base  BC  of  the  triangle 
ABC  meet  the  other  two  sides  at  D  and  E. 

It  is  required  to  prove  that  AD :  DB  =  AE :  EC. 

Proof.  There  are  two  cases  to  be  considered,  according  as 
AD  and  DB  are  commensurable  or  incommensurable. 

Case  I.   When  AD  and  DB  are  commensurable. 

Let  AD  and  DB  have  a  common  measure  which  is  contained 
m  times  in  AD,  and  n  times  in  DB. 

Then  AD  :  DB  =  m  :  n. 

Apply  this  measure  to  AD  and  DB,  and  through  the  points  of 
division  draw  lines  parallel  to  BC,  producing  them  to  cut  the 
side  AC. 

The  number  of  segments  thus  formed  in  ^^and  ECis  equal 
to  the  number  in  AD  and  DB,  respectively,  namely,  pi  seg- 
ments in  AE,  and  n  in  EC. 

Through  the  several  points  of  division  in  the  side  AC  draw 
lines  parallel  to  AB.    These  are  all  equal.     Why  ? 


242]  SIMILAR  RECTILINEAR  FIGURES  157 

Therefore  the  small  triangles  in  the  figure  are  identically- 
equal,  and  the  segments  of  AC  are  all  equal,         (Arts.  94,  43.) 

consequently  AE  :  EC  =  m:n. 

But  AD:DB  =  m:n. 

Therefore  AD:I)B  =  AE:  EC.  (Axiom  1.) 

Case  II.     When  AD  and  DB  are  incommensurable. 


Take  any  measure  of  AD  for  unit,  and  with  it  mark  off 
segments  on  AD  and  DB. 

Since  AD  and  DB  are  incommensurable,  this  unit  will  not 
measure  DB  integrally,  but  will  leave  a  remainder  less  than 
the  unit,  say  B'B. 

Through  B'  draw  a  line  parallel  to  BC,  or  to  DE,  meeting 
AC  at  C. 

Then  in  AAB'O,  since  AD  and  DB'  are  commensurable, 

AD      AE  ,     ^       -r 

DB=EC''  ^^^'''^' 

If  now  the  length  of  the  unit  of  measure  be  repeatedly 
diminished,  B'  will  approach  more  and  more  nearly  to  B, 
and  C  more  and  more  nearly  to  C,  since  B'B  is  always  less 
than  the  unit  of  measure;  and  in  this  way  B'B  can  be  made 
less  than  any  assigned  quantity. 

AD 

Therefore  the  ratio   — —   will   approach  as   its   limit  the 

,.     AD  DB' 

ratio  — ; 


158  ELEMENTARY  GEOMETRY  [Chap.  Ill 

And  the  ratio  — — -  will  approach  as  its  limit  the  ratio 

EC  EG 

A.D  A.E 

But  these  ratios,  '— —  and  — — ,  are  always  equal  to  each 
UIj  eo 

other. 

Therefore  their  limits,  —  and  ^  are  equal.    (Art.  230.) 

That  is,  AD:DB  =  AE:  EC. 

Note.  The  theorem  of  this  proposition  might  be  stated :  A  straight  line 
parallel  to  one  side  of  a  triangle  cuts  the  other  tioo  sides  proportionally. 

243.  Corollary.     In  the  diagram  for  this  proposition 

AB:DB  =  AC:EC',  (Art.  237.) 

AD:AB  =  AE:AC;  (Art.  237.) 

AD  :  AE  =  AB:  AC',  (Art.  234.) 
etc. 

244.  Converse  of  Proposition  I.  If  a  straight  line  divides 
two  sides  of  a  triangle  in  the  same  ratio,  it  is  parallel  to  the 
third  side. 

A 


Let  the  straight  line  DE  divide  the  sides  AB  and  AC  of  the 
triangle  ABC  in  the  same  ratio,  so  that 

AB:AD  =  AC:AE. 


242-244]  SIMILAR  RECTILINEAR  FIGURES  159 

If  DE  is  not  parallel  to  BC,  draw  DE'  parallel  to  BG. 

Then  AB:AD  =  AC:  AE'.        (Prop.  I,  Cor.) 

But  by  hypothesis   AB:AD  =  AC:  AE. 

Therefore  AE'  =  AE  (Art.  238),  so  that  the  lines  DE  and 
DE'  coincide. 

Therefore  DE  is  parallel  to  BG. 

EXERCISES 

1.  If  two  sides  of  a  quadrilateral  are  parallel,  any  straight  line  drawn 
parallel  to  them  will  cut  the  other  two  sides,  or  these  two  sides  produced, 
proportionally. 

2.  ABC  is  a  triangle,  and  through  D,  a  point  in  AB,  DE  is  drawn 
parallel  to  BC,  meeting  AC  in  E.  Through  C  a  straight  line  CF  is 
drawn  parallel  to  BE,  meeting  AB  produced  in  F.  Prove  that  AB  is 
a  mean  proportional  between  AD  and  AF,  i.e.  AD  :  AB  =  AB  :  AF. 

3.  The  straight  line  which  joins  the  mid-points  of  two  sides  of  a 
triangle  is  parallel  to  the  third  side. 

4.  The  straight  line  drawn  through  the  mid-point  of  one  side  of  a 
triangle  parallel  to  a  second  side  bisects  the  third  side. 

5.  Any  two  straight  lines  cut  by  any  three  parallel  straight  lines  are 
cut  proportionally. 

6.  BAC  is  any  angle  and  P  any  point  within  it.  Through  P  draw 
a  straight  line  terminated  both  ways  by  the  boundaries  of  the  angle 
which  is  bisected  at  P. 

Suggestion.  Through  Pdraw  PD  parallel  to  BA  meeting  AC  in  D. 
Find  E  in  AC  so  that  AD  =  DE.  Join  EP  and  produce  to  meet  AB 
in  F.     EF  is  the  required  line. 

7.  BAC  is  any  angle  and  P  any  point  within  it.  Through  P  draw 
a  straight  line  terminated  both  ways  by  the  boundaries  of  the  angle 
and  divided  in  a  given  ratio  at  P. 

Suggestion.  Notice  the  difference  between  this  problem  and  Ex.  6. 
Follow  the  construction  there  given,  making  changes  corresponding  to  the 
changes  in  the  problem. 

8.  If  a  system  of  parallel  lines  intercept  equal  segments  on  one  trans- 
versal, they  will  intercept  equal  segments  on  every  other  transversal. 


160  ELEMENTARY  GEOMETRY  [Chap.  Ill 


Proposition  II 

245.  The  bisector  of  an  angle  of  a  triangle  divides  the 
opposite  side  into  segments  proportional  to  the  adjacent 
sides. 


Let  AD  bisect  the  angle  A  of  the  triangle  ABC,  meeting  the 
opposite  side  at  D. 

It  is  required  to  prove  that  BD  :  DC  =  BA  :  AC. 

Proof.     Through  B  draw  a  line  parallel  to  DA,  meeting  CA 
produced  at  E. 


Then 

ZABE  =  Z.BAD. 
=  Z  CAD. 

Why? 

=  Z  AEB. 

Why? 

Therefore 

AE  =  AB. 

(Art.  51.) 

Now 

BD:DC=EA:  AC. 

(Prop.  I.) 

Therefore 

BD:DC=BA:  AC. 

EXERCISES 

1.  State  and  prove  the  converse  of  the  theorem  in  Proposition  II. 

2.  Apply  this  theorem  to  prove  that  the  straight  line  bisecting  the 
vertical  angle  of  an  isosceles  triangle  also  bisects  the  base. 

3.  ABC  is  any  triangle  whose  base  is  bisected  at  D  ;  the  Zs  ADB  and 
ADC  are  bisected  by  straight  lines  meeting  the  sides  AB  and  AC  at  E 
and  F,  respectively.     Prove  that  EF  is  parallel  to  BC. 


245-248]  SIMILAR  RECTILINEAR   FIGURES  161 

4.    Trisect  a  given  line-segment. 

Suggestion.  Construct  a  triangle  having  the  given  line-segment  for 
base,  and  one  of  the  sides  double  of  the  other.     Bisect  the  vertical  angle. 

Definitions 

246.  If  AB  is  a  given  line-segment  and  C  any  point  between 
A  and  B,  C  is  said  to  divide  AB  internally  into  the  two  seg- 
ments AC  and  CB.     But 

if  C  lies  on  the  given  line    , £ . £ 

A  B 

outside   the  segment  AB, 

as  at  C",  then  it   is  said  to    divide   AB  externally   into   the 

segments  AC  and  OB. 

The  two  segments  are  the  intercepts  between  the  point  and 
the  extremities  of  the  given  line-segment,  and  their  sum,  proper 
attention  being  paid  to  signs,  is  always  equal  to  the  given  line- 
segment.     See  Article  204. 

With  this  definition  the  following  theorem  may  be  stated 
and  proved  similarly  to  that  of  Proposition  II. 

Theorem.  The  bisector  of  an  external  ayigle  of  a  triangle 
divides  the  opposite  side  externally  into  segments  proportional 
to  the  adjacent  side;  and  conversely. 

247.  If  the  angles  of  one  rectilinear  figure  taken  in  order 
are  equal,  respectively,  to  the  angles  of  another  rectilinear 
figure  taken  in  order,  the  two  figures  are  said  to  be  mutually- 
equiangular. 

248.  Two  polygons  are  said  to  be  similar  when  they  have 
the  same  number  of  sides,  are  mutually  equiangular,  and  have 
their  pairs  of  corresponding  sides  proportional. 

Two  polygons  of  the  same  number  of  sides  may  in  general  be  mutually 
equiangular  without  having  their  corresponding  sides  proportional ;  or, 
they  may  have  their  corresponding  sides  proportional  and  not  be  mutually 
equiangular.  Such  polygons  are  not  similar  since  they  do  not  fulfil  all 
the  conditions  of  similarity. 

Similar  polygons  are,  so  to  speak,  of  the  same  shape  but  not  necessarily 
of  the  same  size.  « 


162  ELEMENTARY  GEOMETRY  [Chap.  Ill 


Proposition  III 

249.  Two  triangles  are  similar  if  the  three  angles  of 
one  are  equal,  respectively,  to  the  three  angles  of  the 
other. 


Let  ABC  and  A'B'C  be  two  triangles  in  which  the  angle  A 
equals  the  angle  A',  B  equals  B'  and  C  equals  C 

It  is  required  to  prove  that  AB  :  BC  =  A'B' :  B'C'y 

BO:  CA  =  B'C':  C'A', 

CA:AB=CA':A'B', 

or,  more  briefly,  that       AB  :  BC :  CA  =  A'B' :  B'C  :  C'A'. 

[Will  this  prove  the  triangles  similar  ?    Recall  the  definition  of 
similar  polygons.] 

Proof.  Place  A  A'B'C  upon  A  ABC  so  that  the  vertex  B' 
coincides  with  the  vertex  B  and  Z  A'B'C  with  Z  ABC. 

Then  A'  will  fall  on  BA  or  BA  produced,  and  C  will  fall  on 
BC  or  BC  produced,  and  A'C  will  be  parallel  to  AC     Why  ? 

Therefore  AB:BC=  A'B' :  B'C.  (Prop.  I.) 

Similarly  the  other  pairs  of  sides  may  be  shown  proportional, 
two  and  two. 

250.  Corollary.  If  two  triangles  have  two  angles  of  one 
equal,  respectively,  to  two  angles  of  the  other,  the  triangles  are 
similar. 


249-251]  SIMILAR  RECTILINEAR  FIGURES  163 


Proposition  IV 

251.  If  two  triangles  have  an  angle  of  one  equal  to  an 
angle  of  the  other,  and  the  sides  including  these  angles 
proportional,  the  triangles  are  similar. 


B  C  B' 

Let  ABC  and  A'B'C  be  two  triangles  having  the  angle  at  A 
equal  to  the  angle  at  A',  and  the  sides  AB  and  AC  propor- 
tional to  AB'  and  A'O,  that  is,  having 

AB:AC=AB':A'C'. 

It  is  required  to  prove  that  the  two  triangles  are  similar. 

What  do  we  need  to  show  in  order  to  prove  the  two  triangles 
similar  ? 

Proof.  Suggestions.  (1)  Superpose  the.  triangles  so  that 
the  equal  angles  coincide. 

(2)  The  sides  B'C  and  BC  are  then  parallel.     Why  ? 

(3)  The  triangles  are  therefore  mutually  equiangular  and 
hence  similar.  (Prop.  III.) 

EXERCISES 

1.  Two  right  triangles  are  similar  if  an  acute  angle  of  the  one  equals 
an  acute  angle  of  the  other. 

2.  Two  isosceles  triangles  are  similar  if  their  vertical  angles  are  equal. 

3.  Two  chords  A  C  and  BD  of  a  circle  ABC  intersect  at  E,  either  within 
or  without  the  circle  ;  prove  AsAEB  and  CED  similar,  and  also  /SsAED 
and  BEC. 

4.  The  straight  line  joining  the  mid-points  of  two  sides  of  a  triangle 
is  equal  to  half  of  the  third  side. 


164 


ELEMENTARY  GEOMETRY 


[Chap.  Ill 


252.  Corollary  to  Proposition  IV.  If  two  similar  poly- 
gons are  divided  into  triangles  by  the  diagonals  drawn  from  a 
pair  of  liomologous  vertices,  these  triangles  are  similar  each  to 
each  and  are  similarly  placed. 


Suggestions  for  Proof. 

First.    By  hypothesis  polygons  ABODE  and  A'B'C'D'E'  are  similar. 
That  is  ZB  =  ZB'  and  AB:BC=  A'B> :  B'C. 
Therefore  As  ABC  and  A' B'C  are  similar. 

Next.  ZACB  =  ZA' CD'.  Why  ? 

AC:CD  =  A'C:  CD'.  Why  ? 

Therefore  As  ACD  and  A' CD'  are  similar.     And  so  on. 

Conversely.  If  the  diagonals  drawn  from  one  vertex  in 
each  of  two  polygons  divide  them  into  the  same  number  of  tri- 
angles, similar  each  to  each  and  similarly  placed^  the  two  polygons 
are  similar. 

EXERCISES 

1.  State  the  theorem  in  the  first  chapter  analogous  to  Proposition  IV, 
proving  the  equality  of  two  triangles. 

2.  From  a  point  E  in  the  common  base  of  the  triangles  ACB  and  ADB 
straight  lines  are  drawn  parallel  to  ^C  and  AD  meeting  BC  and  BD  in 
F  and  G  ;  show  that  FG  and  CD  are  parallel. 

3.  If  two  sides  of  a  quadrilateral  are  parallel,  the  diagonals  cut  each 
other  in  the  same  ratio. 

4.  The  bisector  of  an  external  angle  of  a  triangle  divides  the  opposite 
side  externally  into  segments  proportional  to  the  adjacent  sides,  and 
conversely. 


252-253]  SIMILAR  RECTILINEAR  FIGURES  165 


Proposition  V 

253.  If  the  ratios  of  the  three  sides  of  one  triangle  to 
the  three  sides  of  another,  two  and  two,  are  equal,  the 
triangles  are  similar. 
A    ' 


B  OB'  O' 

Let  ABC  and  A'B'C  be  two  triangles  such  that 
AB  :  A'B'  =  AC:A'C'  =  BC :  B'C. 
It  is  required  to  prove  that  the  two  triangles  are  similar. 

What  must  be  shown  in  order  to  prove  the  triangles  similar  ? 
Proof.     On  AB  mark  off  a  length  AD  equal  to  A'B',  and  on 
AC,  a  length  AE  equal  to  A'C     Join  DE. 
Then  because  AB  :  AD  =  AC :  AE 

A  ADE  is  similar  to  A  ABC;  (Prop.  TV.) 

Therefore  AB  :  AD  =  BC :  DE. 

But,  by  hypothesis,  AB  :  A'B'  =  BC :  B'C, 
and  by  construction,  AD  =  A'B'. 

Therefore  DE  =  B'C.  (Art.  238.) 

Hence  A  ADE  is  identically  equal  to  A  A'B'C.       (Art.  53.) 
But  A  ADE  is  similar  to  A  ABC. 
Therefore  A  A'B'C  is  similar  to  A  ABC. 

EXERCISES 

1.  State  the  analogous  proposition  on  the  equality  of  two  triangles. 

2.  Two  triangles  are  similar  when  their  sides  are  respectively  parallel 
or  respectively  perpendicular  to  each  other, 

3.  If  any  quantities  A,  B,  C.  and  p,  q,  r,  are  so  related  that  A  :  C  =p  :  r 
and  B  :  C  =  q  :  r,  show  that  then  A  +  B  :  G  =  p  +  q  :  r. 


166  ELEMENTARY  GEOMETRY  [Chap.  Ill 

Proposition  VI 

254.  The  periineters  of  two  similar  polygons  are  in  the 
same  ratio  as  any  two  corresponding  sides. 

Suggestions  for  Proof. 

1.  Let  ABODE  ...  and  A'B'C'D'E'  •"  be  any  two  similar 
polygons. 

2.  By  definition  AB  :  A'B'  =  BC:B'C'=CD:  CD'  etc. 

3.  Applying  Art.  240, 

AB -}- BC -{- CD  +  ... :  A'B'  +  B'C  +  CD' ...  =  AB.A'B'. 
Therefore 

Proposition  VII 

255.  If  in  a  right  triangle  the  perpendicular  is  drawn 
from  the  vertex  of  the  right  angle  to  the  hypotenuse : 

(1)  The  two  triangles  thus  formed  are  similar  to  each 
other,  and  to  the  whole  triangle; 

(2)  The  perpendicular  is  a  mean  proportional  between 
the  segments  of  the  hypotenuse ; 

(3)  Each  side  of  the  triangle  is  a  mea^n  proportional 
between  the  hypotenuse  and  the  segm^ent  adjacent  to  that 
side. 

C 


Let  ABC  be  a  right  triangle,  right-angled  at  C,  and  CD  the 
perpendicular  from  C  to  AB. 

It  is  required  to  prove 

(1)  That  the  triangles  ACD  and  CBD  are  similar  to  each 
other  and  to  ABC. 

All  these  triangles  are  easily  shown  to  be  mutually  equi- 
angular and  therefore  similar  by  Proposition  III. 


254-257]  SIMILAR  RECTILINEAR   FIGURES  167 

(2)  That  AD:  DC  =  DC:  DB. 

This  follows  immediately  from  the  fact  that  As  ADC  and 
CDB  are  similar. 

(3)  That  AB'.AG=AC:AD, 
and                               AB:BC  =  BC:BD. 

These  relations  follow  from  the  fact  that  each  small  triangle  is 
similar  to  the  whole  triangle. 

256.  Corollary.  A  perpendicular  drawn  from  any  point 
of  a  circle  to  a  diameter  is  a  mean  proportional  between  the 
segments  into  which  it  divides  the  diameter. 

EXERCISES 

1.  Show  that  the  radius  of  a  circle  is  a  mean  proportional  between  the 
segments  of  a  tangent  intercepted  between  its  point  of  contact  and  any 
pair  of  parallel  tangents. 

2.  If  in  two  similar  triangles  perpendiculars  are  drawn  from  homolo- 
gous vertices  to  the  opposite  sides,  these  are  in  the  same  ratio  as  a,ny  pair 
of  homologous  sides. 

3.  The  sides  of  a  triangle  ABC  are  6,  7,  and  8  feet,  respectively  ;  find 
the  lengths  of  the  segments  into  which  the  bisector  of  the  opposite  angle 
will  divide  each  side. 

4.  Show  that  the  diagonals  of  any  quadrilateral  inscribed  in  a  circle 
divide  the  quadrilateral  into  four  triangles,  which  are  similar,  two  and 
two. 

5.  Sum  up  the  conditions  under  which  two  triangles  are  similar  and 
compare  these  with  the  conditions  under  which  two  triangles  are  identi- 
cally equal  as  given  in  Chapter  I. 

*257.  We  shall  hereafter  frequently  find  it  convenient  to 
speak  of  the  product  of  two  line-segments,  by  which  we  shall 
mean  the  product  of  their  measures  by  a  common  unit.  With 
this  understanding  we  may  enunciate  the  following  two 
theorems. 


168  ELEMENTARY  GEOMETRY  [Chap.  Ill 


Proposition  VIII 

258.  If  two  chords  of  a  circle  intersect  within  the  circle 
the  product  of  the  segments  of  the  one  equals  the  product 
of  the  segments  of  the  other. 


Let  AB  and  CD  be  two  chords  intersecting  at  S  within  a 
given  circle. 

It  is  required  to  prove  that  the  product  of  the  segments  AS 
and  SB  equals  the  product  of  the  segments  CS  and  SD. 

Proof.     Join  AD  and  BC. 

As  ASD  and  CSB  are  mutually  equiangular.  (Art.  181.) 

Therefore  these  triangles  are  similar,  (Prop.  III.) 

and  AS:SD=CS:  SB. 

Therefore  the  product  AS  -  SB  =  the  product  CS  -  SD. 

(Art.  233.) 
This  theorem  might  be  stated : 

If  two  cJiords  intersect  within  a  circle,  their  segments  are 
recipi'ocally  proportional 

A  segment  of  the  first  chord  is  to  a  segment  of  the  second  as  the  re- 
maining segment  of  the  second  is  to  the  remaining  segment  of  the  first; 
that  is  the  segments  are  taken  in  reciprocal  orders  in  the  two  ratios. 

259.  Corollary  I.  If  through  a  fixed  point  within  a  circle 
any  chord  is  di'aivn,  the  product  of  its  segments  is  the  same 
whatever  its  direction. 


258-262]  SIMILAR  RECTILINEAR  FIGURES  169 

260.  Corollary  II.  Either  segment  of  the  least  chord  that 
can  he  drawn  through  a  fixed  point  within  a  circle  is  a  mean  pro- 
portional between  the  segments  of  any  other  chord  draivn  through 
that  point. 

Proposition  IX 

261.  If  two  chords  of  a  circle,  when  produced,  intersect 
without  the  circle,  the  product  of  the  seginents  of  the  one 
equals  the  product  of  the  segments  of  the  other. 


Let  AB  and  CD  be  two  chords  intersecting  at  S  without  the 
circle. 

It  is  required  to  prove  that  the  product  of  the  segments  AS 
and  SB  equals  the  product  of  the  segments  CS  and  SD. 

Proof.     Join  AD  and  BC. 

The  proof  is  identical  with  that  of  Proposition  VIII. 

• 
262.   Corollary  I.     If  a  tangent  and  a  secant  of  a  circle 
intersect,  the  tangent  is  a  mean  proportional  between  the  whole 
secant  and  its  external  segment. 

This  follows  directly  from  the  preceding  T 

proposition  on  considering  a  tangent  as  the 
limit  of  a  secant,  i.e.  as  a  secant  whose 
points  of  intersection  coincide.  Or  it  may 
easily  be  proved  after  the  same  general 
method  as  was  used  in  the  two  preceding 
propositions,  joining  TA  and  TB. 


170  ELEMENTARY  GEOMETRY  [Chap.  Ill 

263.  Corollary  II.  If  from  any  point  on  the  common  chord 
of  two  intersecting  circles  produced,  tangents  are  drawn  to  the 
circles,  the  lengths  of  these  tangents  are  equal. 

For  either  tangent  is  a  mean  proportional  between  the  seg- 
ments into  which  the  chord  is  divided  externally  by  the 
point. 

Definition".  The  common  chord  of  two  intersecting  circles 
produced  indefinitely  is  called  the  Radical  Axis  of  the  circles ; 
and  a  system  of  circles  through  the  same  two  points  is  called 
a  coaxial  system,  since  all  pairs  of  the  circles  have  the  same 
radical  axis. 

264.  It  will  be  interesting  to  recall  the  principle  of  continuity  in 
connection  with  Propositions  VIII  and  IX. 

Proposition   VIII   states  that  if   two  chords  AB  and   CD  intersect 
within  a  circle  at  S,  then  AS  •  SB 
=  CS  .  SD.  ^/ 

Named  in  that  order  the  two        X-^^ 
segments  of  each  chord  are  of  the  ^N^"*^^ 

same  sign. 

Suppose,  now  we  rotate  the 
chord  CD  about  the  point  D. 
Then  when  C  coincides  with  A, 
S  will  also  coincide  with  A  and 
one  segment  of  each  chord  will 
vanish.  Both  products  AS  •  SB 
and  CS  •  SD  then  become  zero 
and  are  still  equal. 

If  we  continue  the  rotation,  S  will  pass  outside  the  circle,  i.e.  the  two 
chords  will  intersect  outside  the  circle,  as  in  Proposition  IX,  and  the 
point  of  intersection  will  divide  them  both  externally.  The  two  segments 
AS  and  SB  will  then  be  of  opposite  signs,  as  will  also  CS  and  SD,  and 
the  products  AS  ■  SB  and  CS  •  SD  still  remain  equal. 

These  two  propositions  may  thus  be  comprised  under  a  single  more 
general  theorem,  as  follows : 

The  products  of  the  segments  of  any  two  chords  of  a  circle.^ 
determined  by  their  point  of  intersection,  are  equal. 


263-265]  SIMILAR   RECTILINEAR  FIGURES  171 


EXERCISES 

1.  If  two  line-segments  intersect  so  as  to  make  the  product  of  the  seg- 
ments of  one  equal  to  the  product  of  the  segments  of  the  other,  their  four 
extremities  are  concyclic. 

Suggestion.  Pass  a  circle  through  three  of  their  extremities,  and 
apply  Proposition  VIII. 

2.  If  two  circles  intersect  and  through  any  point  of  their  common 
chord  two  other  chords  are  drawn,  one  in  eacji  circle,  their  four  extremi- 
ties are  concyclic. 

3.  If  two  circles  intersect,  their  common  chord  bisects  their  common 
tangents. 

4.  If  two  circles  touch  each  other,  the  tangent  at  their  point  of  contact 
bisects  their  other  common  tangents. 

6.  If  three  circles  intersect  two  and  two,  their  three  common  chords 
pass  through  one  point. 

SuGGESTiox.  If  AB  and  CD,  the  common  chords  of  X,  Y,  and  X,  Z, 
intersect  at  0,  and  P,  one  of  the  common  points  of  Y  and  Z,  is  joined  to 
0,  then  PO  must  pass  through  the  other  common  point  of  Y  and  Z.  If 
not  let  it  cut  Y  a  second  time  at  Q,  and  Z  a  second  time  at  R,  and  prove 
OQ  =  OR. 

6.  AB  is  a  given  line-segment,  D  and  E  two  points  on  it ;  DF  and  EG 
are  parallel  and  proportional  to  AD  and  AE.  Prove  that  Aj  F,  and  G 
lie  on  one  straight  line. 

7.  The  line-segment  AB  is  divided  internally  at  C  and  Z>,  so  that 
AB  :  AG  =  AC  :  AD.  From  A  any  other  line-segment  AE  is  drawn  equal 
to  AC.  Prove  that  the  triangles  ABE  and  AED  are  similar,  and  that 
EC  bisects  the  angle  BED. 

265.  We  shall  here  insert  a  proposition  which  has  no  direct 
connection  Avith  the  properties  considered  in  this  chapter,  but 
which  will  serve  as  an  additional  example  of  a  demonstration 
by  the  Method  of  Limits,  and  recall  some  of  our  fundamental 
notions  of  Ratio.  It  will  be  well  for  the  pupil  to  review  the 
proof  of  Proposition  I  in  this  connection.  These  two  propo- 
sitions are  of  the  same  general  character,  and  the  method 
used  in  demonstrating  them  will  be  frequently  used  in  subse- 
quent demonstrations. 


172  ELEMENTARY  GEOMETRY  [Chap.  Ill 


Proposition  X 

266.   I^^  equal  circles,  angles  at  the  centre  are  in  the 
same  ratio  as  the  arcs  subtending  them. 


Let  AOB  and  A'O'B'  be  angles  at  the  centres  of  two  equal 
circles,  subtended  by  the  arcs  AB  and  A'B',  respectively. 

It  is  required  to  prove 

Z  AOB  :  Z  A'O'B'  =  arc  AB  :  arc  A'B'. 

Proof.     Two  cases  arise  according  as  the  arcs  AB  and  A'B' 
are  commensurable  or  not. 

Case  I.     Suppose  AB  and  A'B'  to  have  a  common  measure, 
which   is    contained   m   times   in  AB  and  n   times  in  A'B'. 
Apply  this  measure  to  the  arcs  and  mark  the  points  of  division. 
Draw  radii  to  the  points  of  division. 

These  radii  divide  Z.AOB  into  m  equal  parts  and  Z  A'O'B' 
into  n  parts,  equal  to  each  other  and  to  the  parts  of  Z  AOB. 

(Arts.  159,  160.) 
arc  AB      m 


Then 
and 

Therefore 


arc  A'B'      n 
ZAOB       m 


Z  A'O'B'      n 
ZAOB        arc  ^5 


Z  A'O'B'      arc  A'B' 
or  Z  AOB :  Z  AO'B'  =  arc  AB :  arc  A'B'. 


266-268]  SIMILAR  RECTILINEAR   FIGURES  173 

Case  II.  Suppose  AB  and  A'B'  to  be  incommensurable. 
Take  any  measure  of  A'B',  say  x,  and  apply  it  to  AB  as  often 
as  possible.  Let  the  last  point  of  division  be  C,  so  that  the 
arc  CB  is  less  than  x. 


Then  the  arcs  AG  and  A'B'  are  commensurable,  so  that 

Z^OC^arc^O^  by  Case  I. 

ZA'O'B'     SiVG  A'B'  ^ 

By  repeatedly  decreasing  the  unit  measure  x,  the  arc  CB 
cau  be  made  smaller  than  any  assigned  quantity,  so  that  the 
arc  AC  will  approach  as  its  limit  the  arc  AB,  and  at  the  same 
time  Z  AOG  will  approach  as  its  limit  Z  AOB. 

Therefore  the  ratios  ^^    and  f"    ,  approach  as  their 

ZA'O'B'         3i,vG  A'B'    ^^ 

limits  the  ratios  — — - — —-  and — -,  respectively. 

ZA'O'B'  3bVG  A'B' 

But    ^^^^  =  avG  AC   ^  ^^^.^  measure  x. 

ZA'O'B'      ^vcA'B'  ^ 

Therefore  the  limit    ^^^^  =  the  limit  _?^£^_4^. 
ZA'O'B'  SiVGA'B' 

(Art.  230.) 

Or  Z  AOB  :  Z  A'O'B'  =  arc  AB  :  arc  A'B'. 

267.  Corollary  I.  In  the  same  circle,  angles  at  the  centre 
are  in  the  same  ratio  as  the  arcs  subtending  them. 

268.  Corollary  II.  In  the  same  circle,  or  in  equal  circles, 
angles  whose  vertices  lie  on  the  circle  are  in  the  same  ratio  as  the 
arcs  subtending  them. 


174  ELEMENTARY  GEOMETRY  [Chap.  Ill 

Section    III 

PROBLEMS    OF    CONSTRUCTION 

Proposition  XI 

269.    To   divide   a  given    line-segment    into    any    re- 
quired number  of  equal  parts* 


It  is  required  to  divide  a  giVen  line-segment  AB  into  say  n 
equal  parts. 

Construction.  From  one  end  of  the  given  line-segment  draw 
any  straight  line  AC  making  an  angle  with  AB. 

On  AG  mark  off  n  equal  parts  of  any  convenient  length,  be- 
ginning at  the  point  A,  and  call  the  points  of  division  P,  Q,  R,  ••-, 
the  last  one  being  X. 

Join  XB  and  through  the  points  of  division  P,  Q,  R,  •  •  •  draw 
lines  parallel  to  XB,  cutting  AB  in  the  points  P^,  Qi,  7?i,  •••. 

Then  AB  is  divided  into  n  equal  parts  AP^,  l\Qi,  QiRi,  •••• 

Proof.  Since  PPi,  QQi,  BRi,  -••,  XB  are  parallel  and  inter- 
cept equal  segments  on  AC,  they  likewise  intercept  equal 
segments  on  AB.  (Prop.  I.) 

EXERCISES 

1.  Apply  the  method  of  this  proposition  to  bisect  a  given  line-segment, 
to  trisect  a  given  line-segment. 

2.  Show  how  to  find  three-fifths  of  a  given  line-segment 


269-271]  SIMILAR   RECTILINEAR   FIGURES  175 

Proposition  XII 

270.   To  divide  a  given  line-segment  into  parts  pro- 
portional to  other  given  line-seginents. 


It  is  required  to  divide  a  given  line-segment  AB  into  parts 
proportional  to  the  given  segments  a,  h,  c. 

The  construction  is  similar  to  that  for  Proposition  XI,  except 
that  on  the  arbitrary  line  AC,  segments  equal  to  a,  b,  c,  re- 
spectively, are  marked  off,  instead  of  segments  equal  to  each 

other. 

The  pupil  should  write  out  the  construction  and  proof  in  full. 

271.  If  it  is  required  to  divide  a  given  line-segment  AB  in 
the  ratio  of  a  to  b,  we  must  know  whether  the  division  is  to  be 
internal  or  external.  If  internal,  proceed  just  as  in  the  propo- 
sition above;  but  if  external,  mark  on  the  line  AC  the  segment 
AP  equal  to  a,  and  PQ  equal  to  b  as  before,  but  instead  of  PQ 
being  an  extension  of  AP,  it  should  lie  upon  AP  so  that  AQ  is 
divided  externally  at  P  in  the  given  ratio.  Then  join  QB  and 
draw  PM  parallel  to  QB.  M  will  be  the  required  point  of 
division. 

EXERCISES 

1.  Divide  a  given  line-segment  eight  inches  long  into  three  parts  pro- 
portional to  one,  two,  and  three. 

2.  Divide  a  given  line-segment  eight  inches  long  (1)  internally  in  the 
ratio  3:2;  (2)  externally  in  the  ratio  3  : 2. 


176  ELEMENTARY  GEOMETRY  [Chap.  Ill 


Proposition  XIII 

272.   To  find  the  fourth  proportional  to  three  given 
line-segments. 


1 

^ 

b 

jy^^^ 

\ 

\ 

It  is  required  to  find  the  fourth  proportional  to  the  three 
given  line-segments,  a,  b,  c. 

Construction.  Draw  two  straight  lines  AB  and  AC,  making 
any  angle  with  each  other. 

Upon  AB  mark  off  the  segments  AD  and  DE,  equal  respec- 
tively to  a  and  b. 

Upon  AC  mark  off  the  segment  AF  equal  to  c. 

Join  DF,  and  through  E  draw  EH  parallel  to  DF. 

Then  FH  is  the  fourth  proportional  to  a,  b,  c. 

The  proof  is  apparent  from  Proposition  I. 

It  should  be  noticed  that  if  a,  6,  c  are  taken  in  a  different 
order  the  fourth  proportional  will  be  different. 


EXERCISES 

1.  To  find  the  third  proportional  to  two  given  line-segments. 
What  is  meant  by  a  third  proportional  to  two  quantities  ? 

2.  How  many  different  fourth  proportionals  can  be  found  with  three 
given  line-segments  ? 

3.  AB  and  AC  are  two  straight  lines  drawn  from  A.  Produce  CA 
backward  to  Z>,  making  AD  =  AG;  describe  a  circle  through  the  three 
points  B,  C,  Z),  and  produce  BA  to  meet  this  circle  at  E.  AE  is  a  third 
proportional  to  AB  and  AG. 


272-274]  SIMILAR   RECTILINEAR  FIGURES  177 


Proposition  XIV 

273.   To    find    the    mean    proportional    between    two 
given  line-segments. 


It  is  required  to  find  the  mean  proportional  between  the  two 
given  line-segments  a  and  h. 

Construction.  Upon  any  straight  line  mark  off  the  segments 
AB  and  BC  equal  to  a  and  h,  respectively. 

Upon  ^O  as  diameter  describe  a  circle,  and  at  B  erect  a  per- 
pendicular to  AC,  meeting  the  circle  at  D. 

Then  BD  is  the  mean  proportional  between  a  and  c. 

(Art.  256.) 
EXERCISES 

1.  If  the  given  line-segments  were  ^Cand  BC,  placed  as  in  the  dia- 
gram, show  how  to  find  a  mean  proportional  between  them. 

2.  Show  that  half  the  sura  of  two  unequal  line-segments  is  greater 
than  the  mean  proportional  between  them. 

\' 

274.  Definition.  When  a  given  line-segment  is  divided 
into  two  parts  such  that  one  of  the  parts  is  a  mean  proportional 
between  the  whole  segment  and  the  other  part,  it  is  said  to  be 
divided  in  extreme  and  mean  ratio. 

For  example,  if  AB  is  divided  internally        • -t, :• 

at  C  so  that  ^  ^ 

AB  :  AC  =  AC :  CB, 

then  AB  is  divided  internally  in  extreme  and  mean  ratio ;  while  if  C  is  so 
chosen  on  AB  produced  that 

AB:BC=BC:AC, 
then  AB  is  divided  externally  in  extreme  and  mean  ratio. 

N 


178  ELEMENTARY  GEOMETRY  [Chap.  Ill 


Proposition  XV 

275.   To  divide  a  given  line-segment  in  extreme  and 
mean  ratio. 


It  is  required  to  divide  the  line-segment  AB  at  a  point  0,  so 
that  AB.AC=AC.CB. 

Construction.  At  B  draw  a  straight  line  BO  perpendicular 
to  AB  and  equal  to  half  of  AB. 

With  centre  0  and  radius  OB  describe  a  circle.  Draw  AO, 
cutting  the  circle  at  D  and  D\ 

On  AB  mark  off  a  segment  AC  equal  to  the  shorter  segment 
AD. 

Then  C  is  the  point  required. 

Proof.     AB  is  tangent  to  the  circle  DBD\  (Art.  187.) 

Therefore  AD'  :AB  =  AB:  AD  or  AC.       (Art.  262.) 

Whence,  AD'  -  AB:  AB  =  AB  -  AC :  AC.  (Art.  237.) 

Now  AB  =  2  0B  =  DD', 

and  AD'  -AB=  AD'  -  DD'  =  AD  =  AC; 

also  AB-AC=BC. 

Therefore  AC:AB  =  BC:  AC, 

or,  by  inversion,  AB  :  AC  =  AC :  BC 

EXERCISE 

1.  Suppose  on  BA  produced  you  should  mark  off  a  segment  ^^  equal 
to  AD'.  Prove  tliat  AB  :  AK=  AK  :  BK^  or  that  AB  is  divided  exter- 
nally at  D  in  extreme  and  mean  ratio. 


275-276]  SIMILAR  RECTILINEAR   FIGURES  179 


Proposition  XVI 

276.  Upon  a  given  line-segment  to  construct  a  polygon 
similar  to  a  given  polygon,  and  such  that  the  given 
line-segment  shall  be  homologous  to  a  given  side. 


It  is  required  to  construct  upon  A^B^  a  polygon  similar  to 
ABODE,  so  that  A'B'  shall  be  homologous  to  the  side  AB. 

Construction.  Divide  the  given  polygon  into  triangles  by 
drawing  the  diagonals  from  the  vertex  A. 

Make  ZsA'B'C  and  B'A'C  equal  to  ZsABC  and  BAC, 
respectively. 

Then  A  A'B'C  is  similar  to  A  ABC.  (Prop.  II.) 

Similarly,  construct 

A  A'C'D'  similar  and  similarly  situated  to  A  ACD, 
and     A  A'D'E'  similar  and  similarly  situated  to  A  ADE. 
Then  the  polygon 
A'B'C'D'E'  is  similar  to  the  polygon  ABODE.    (Art.  252.) 

EXERCISES 

1.  Inscribe  in  a  given  circle  a  triangle  similar  to  a  given  triangle. 
Suggestion.     At  any  point  of  the  circle  draw  the  tangent  and  chord 

making  an  angle  equal  to  one  angle  of  the  given  triangle.     Then  apply 
Article  198. 

2.  Circumscribe  about  a  given  circle  a  triangle  similar  to  a  given 
triangle. 


180  ELEMENTARY  GEOMETRY  [Chap.  Ill 

Section  IV 

ADDITIONAL  PROPOSITIONS 

Proposition  XVII 

277.  //  the  sides  of  a  triangle  are  cut  by  any  straight 
line,  the  product  of  the  ratios  of  their  segments  taken 
in  order  equals  unity. 


Let  the  sides  BC,  CA,  AB,  or  these  sides  produced,  be  cut 
by  a  straight  line  in  the  points  A',  B',  C,  respectively. 

It  is  required  to  prove  that   —  •  —  .  —  =  1. 
^  A'C   B'A    C'B 

If  we  give  heed  to  the  signs  of  the  segments,  this  product  is 
—  1,  since  BA'  and  ^'C  are  of  opposite  signs,  while  CB'  and 
B'A^  AC  and  C'B  are  of  the  same  sign  ;  that  is,  the  first 
ratio  is  negative,  the  second  and  third  positive.  If  the  line 
cuts  all  three  sides  produced,  the  three  ratios  are  all  nega- 
tive, and  their  product  likewise  is  negative. 

Proof.     Draw  perpendiculars  a,  b,  c  from  the  vertices  A,  B,  C, 
of  the  triangle,  to  the  given  transversal. 

Then  M  =  ^.,    CB[^c     A^^ a. 

A'C      c    B'A     a'    C'B      b         ^       ^         ^ 
Therefore 

BA'    CB'    AC^b    c    a^^ 

A' C' B'A    C'B      c' a' b 

This  theorem  is  attributed  to  Menelaus,  a  Greek  geometer 
who  lived  in  the  latter  part  of  the  first  century. 


277-278]  SIMILAR   RECTILINEAR  FIGURES  181 

278.  Conversely.  If  on  the  sides  BC,  CA,  AB  of  a  tri- 
angle ABC,  three  points,  A',  B',  C,  respectively,  be  so  chosen 

BA'    PR'     AC 

as  to  fulfil  the  relation  ^-  ^  *  ^=  ~  ^   (attention  being 

paid  to  the  signs  of  the  segments),  these  three  points  are  collinear. 

Proof.  Let  the  straight  line  A'B^  cut  the  side  AB  of  the 
triangle  at  O/. 

Then  —  •  ^  .  ^^  =  - 1  by  the  direct  theorem. 

AC   B'A    C'B  ^ 

But  it  was  assumed  that 

BA'    CB'    AC  ^     ^ 
A'C '  B'A  '  CB 

Therefore  ^  =  ^, 

whence  C  and  O/  coincide. 

If  attention  were  not  paid  to  the  signs  of  the  segments,  the 
last  proportion  could  be  true  and  C  and  Ci'  not  coincide,  for 
in  that  case  C  and  Ci'  might  divide  the  side  AB  internally 
and  externally  in  the  same  ratio. 

EXERCISES 

1.  Points  E  and  F  are  taken  on  the  sides  AC  and  AB  of  a  triangle 
ABC,  such  that  AE  is  twice  EC  and  BF  is  twice  FA  ;  the  straight  line 
FE  produced  cuts  the  side  BC  bX  D.     Find  the  ratio  of  BD  to  DC. 

2.  If  the  bisectors  of  the  angles  B  and  O  of  a  triangle  meet  the  oppo- 
site sides  at  D  and  E,  and  if  the  straight  line  BE  produced  meet  BC 
produced  at  F,  then  the  external  angle  at  A  is  bisected  by  AF. 

3.  If  a  side  BC  of  a  triangle  ABC  is  bisected  by  a  straight  line  which 
meets  the  sides  AB  and  AC,  produced  if  necessary,  at  D  and  E  respec- 
tively, then  AE  :  EC  =  AD  :  DB. 

4.  Two  straight  lines  intersect  at  A,  and  Z)  is  a  point  between  them. 
Draw  through  D  a  straight  line  such  that  the  segments  of  it  intercepted 
between  D  and  the  given  lines  shall  be  in  a  given  ratio. 

Suggestion.  Join  AD  and  produce  to  E  so  that  AD :  DE  in  the 
given  ratio.  Through  E  draw  a  straight  line  parallel  to  one  of  the  given 
lines,  meeting  the  other  at  H.     Then  H  is  a  point  of  the  required  line. 


182 


ELEMENTARY  GEOMETRY 


[Chap.  Ill 


Proposition  XVIII 

279.  If  two  triangles  ABC  and  A'B'C  are  so  situated 
that  the  straight  lines  joining  the  pairs  of  correspond- 
ing  vertices  A,  A' ;  B,  B' ;  C,  C  are  concurrent,  then  the 
points  of  intersection  of  tlw  three  pairs  of  correspond- 
ing sides  are  collinear. 


Let  the  two  triangles  ABC  and  A'B'C  be  so  situated  that 
the  straight  lines  AA',  BB',  CC  meet  in  0. 

It  is  required  to  prove  that  the  points  of  intersection  x,  y,  z, 
of  the  three  pairs  of  corresponding  sides  BC,  B'C  ]  CA,  C'A' ; 
AB,  A'B',  are  collinear. 

Proof.  In  order  to  prove  vi^hat  is  stated  it  will  be  sufficient 
to  show, 

Az     Bx     Qi^_i 
zB  '  xC  '  yA 

For  if  this  relation  is  satisfied,  x,  y,  z,  points  on  the  sides 
of  the  triangle  ABC,  must  be  collinear.  (Art.  278.) 

First,  if  we  consider  A  OAB  to  be  cut  by  the  transversal 
B'A'z,  we  have 

OA'     Az     BB[ 
A' A  '  zB  '  B'O 


(1) 


=  -1. 


(Prop.  XVII.) 


279-280]  SIMILAR  BECTILINEAE  FIGURES  183 

Similarly,  from  As  OBG  and  OCA  cut  by  the  transversals 
B'C'x  and  A'C'y,  respectively,  we  get 

.^x     qB[     Bx      CO  ^     . 
^  ^    bb'  xC'  CO  ' 

^  ^    A'O      CO     yA 

Multiplying  together  the  left-hand  members  of  equations 
(1),  (2),  and  (3),  and  also  their  right-hand  members,  we  obtain 

Az     Bx      Cy  _     ^ 
zB     xC     yA 

Therefore  x,  y,  z,  are  col  linear. 


280.   Conversely.     If  two  triangles  ABC  and  A'B'C  are  so 

situated  that  the  points  of  intersection,  x,  y,  z,  of  the  three  pairs 
of  corresponding  sides,  BC,  B'C ;  CA,  CA';  AB,  A'B',  are 
coUinear,  then  the  straight  lines  AA',  BB',  CC,  joining  the  pairs 
of  corresponding  vertices,  are  concurrent. 

This  may  be  proved  indirectly  as  follows : 

Let  0  be  the  intersection  of  the  lines  BB^  and  CC;  and 
let  the  straight  line  AO  cut  B'z  at  A^,  which,  so  far  as  we 
know,  may  or  may  not  coincide  with  A\ 

Then  As  ABC  and  A^B'O  fulfil  the  conditions  of  the  direct 
theorem;  therefore  A^C  and  AC  must  intersect  on  the  straight 
line  xz. 

But  AC  cuts  xz  at  y. 

Therefore  A^C  must  pass  through  y,  and  hence  A^  must 
coincide  with  A^, 

This  theorem  and  its  converse  are  known  as  Desargues's  theorems, 
named  from  Gerard  Desargues  (born  at  Lyons,  France,  1593,  died  1662), 
to  whom  they  are  usually  attributed,  though  the  geometric  properties 
involved  in  them  were  known  much  earlier. 


184  ELEMENTARY  GEOMETRY  [Chap.  Ill 


Proposition   XIX 

281.  If  through  any  point  O  straight  lines  are  drawn 
to  the  vertices  A,  B,  C  of  a  triangle,  meeting  the  oppo- 
site sides  at  A',  B' ,  C,  respectively,  then  the  product  of 
the  ratios  of  the  segments  of  the  sides  taken  in  order 
equals  unity,  that  is, 

BA'     CB'    AC  ^. 
A'C  '  B'A  '  CB 


Proof.     Take  A  A  A'C  and  consider  BOB'  a  transversal  of  it. 


Then  4^..^.^  =  _1 


OA'     BC     B'A 
also  from  AAA'B  and  the  transversal  COC  we  get 

AO     A'C    BC  ^      ^ 

OA'  '  CB  '  CA  ' 

Dividing  the  first  of  these  results  by  the  second,  and 
remembering  that  BC  =  —  CB,  A'B  =  —  BA',  etc.,  we  obtain, 

BA'     CB'     AC  ^^ 

A'C  '  B'A  '  CB 

The  converse  of  this  theorem  may  easily  be  proved  in- 
directly. 

In  the  diagram  we  have  chosen  the  point  0  within  the 
triangle  ABC  The  relation  is  equally  true  if  the  point  is 
chosen  outside  the  triangle.  This  should  be  verified  both  for 
a  point  0  obtained  by  passing  out  of  the  triangle  across  a  side, 
and  through  a  vertex. 


281-282]  SIMILAR  RECTILINEAR  FIGURES  185 


Proposition  XX 

282.  To  find  the  locus  of  a  point  whose  distances 
from  two  fixed  points  are  in  a  constant  ratio  differ- 
ent from  unity, 

*A. 


Let  A  and  B  be  the  two  fixed  points,  and  M  any  point, 
such  that  MA :  MB  equals  a  given  fixed  quantity  p. 

It  is  required  to  find  the  locus  of  M. 

Construction.  Join  AB  and  divide  this  line-segment  inter- 
nally at  O  and  externally  at  D  in  the  given  ratio  jh  i-e.  so  that 
AC  :CB  =  AD:DB=p.  (Art.  271.) 

Then  MA  :  3fB  ^  AC :  GB. 

Therefore  MC  bisects  Z  AMB.  (Ex.  1,  p.  160.) 

Also  MA:  MB  =  AD:  DB. 

Therefore  MD  bisects  Z  A^MB.  (Ex.  4,  p.  164.) 

Therefore  Z  CMD  is  a  right  angle. 

Whence  the  locus  of  Jf  is  a  circle  of  which  CD  is  a 
diameter. 

EXERCISES 

1.  What  is  the  locus  of  a  point  equidistant  from  two  fixed  points? 
(See  Art.  57.) 

That  is,  what  is  the  locus  of  a  point  whose  distances  from  two  fixed 
points  are  in  a  constant  ratio,  equal  to  unity  f 

2.  How  would  the  diagram  for  Proposition  XX  change  if  the  given 
constant  ratio  were  made  more  and  more  nearly  equal  to  unity  ?  As  the 
ratio  approaches  unity  how  does  the  point  C  move  ?  How  does  the  point 
D  move  ?    What  about  the  circle  ? 


186  ELEMENTARY  GEOMETRY  [Chap.  Ill 


Harmonic   Division  * 

283.  Definition.  Any  line-segment  which  is  divided  in- 
ternally and  externally  in  the  same  ratio  is  said  to  be  divided 
harmonically. 

For  example,  if  the  seg-     2  C      B  7j 

ment  AB,  as   in  the   last 

proposition,  is  divided  internally  at  C  and  externally  at  D  in 
such  way  that 

AC:CB  =  AD:DB, 

then  AB  is  divided  harmonically  at  0  and  D,  and  the  four 
points  A,  B,  C,  D,  are  said  to  form  a  harmonic  range. 

284.  Theorem.  If  any  line-segment  AB  is  divided  har- 
monically at  C  and  D,  the  line-segment  CD  is  also  divided 
harmonically  at  A  and  B. 

For  if  AC:CB  =  AD:  DB, 

so  also  is  CA.AD=CB:  BD,  (Art.  234.) 

which  shows  that  the  segment  CD  is  divided  at  A  and  B 
internally  and  externally  in  the  same  ratio. 

Thus  in  a  harmonic  range  of  four  points  there  are  two  pairs 
of  points  such  that  each  pair  bears  the  same  relation  to  the 
other  pair. 

The  points  of  each  pair  are  called  ^harmonic  conjugates' 
relative  to  the  points  of  the  other  pair,  and  are  said  to  be  sepa- 
rated harmonically  by  them. 

285.  For  the  sake  of  symmetry  we  have  written  the  ratios 
in  the  form  AC :  CB  =  AD  :  DB,  but  since  the  point  D  divides 
the  segment  AB  externally,  while  C  divides  it  internally,  one 

*  This  section  may  be  omitted  on  a  first  reading. 


283-286]  SIMILAR  RECTILINEAR  FIGURES  187 

of  the  four  segments  AC,  CB,  AD,  DB,  must  be  opposite  in 
sign  from  the  other  three ;  and  so,  if  attention  be  given  to  signs, 
we  should  write  either 

AG:CB  =  AD'.-  DB, 

or  AC:CB=AD:BD. 

Choosing  the  latter  form,  we  may  write 

CB     BD 


AC     AD 

or,  since 

CB  = 

■-AB- 

- AC  and  BD  =  AD-  AB, 

AB-AC     AD-AB 
AC               AD 

Hence 

AB     ^      ^      AB 
AC     ^  =  ^      ad' 

or 

AB     AB^^ 
AC^AD 

Therefore 

11         2 
AC     AD     AB 

(1) 

a  formula  of  great  value  in  the  theory  of  Harmonic  Division. 

286.   Another  important  relation  among  harmonic  points  on 
a  line  can  be  deduced  in  a  similar  way. 

Let  A,  C,  B,  D  he  Si  har- 

monic  range,  A  and  B  being     J J- — J • ^ 

harmonica] ly  separated  by 

C  and  D,  and  let  M  be  the  mid-point  of  the  line-segment  CD. 

Then  AM'BM=CM\ 

Proof.     Since  the  points  A,  C,  B,  D  are  harmonic.  A,  B  and 
C,  D  being  the  pairs  of  conjugates, 

AG^AD 
CB     BD 


188  ELEMENTARY  GEOMETRY  [Chap.  lU 

Moreover,  giving  attention  to  the  signs  of  segments, 

AC=AM^MC, 

CB=CM-{-MB, 

AD=AM+MD, 

BD  =  BM+MD; 
and  since  M  is  the  mid-point  of  CD, 

CM=  MD. 
Substituting  in  the  relation 


we  have 


or 


AG 
CB 

AM-\-MC 

AD 

~-bd' 

AM+MD 

CM+MB 
AM-  CM 

BM  +  ML) 
AM+  CM 

CM-BM     BM+CM 


Whence,  multiplying  by  the  common  denominator  and  col- 
lecting terms, 


AM'  BM=  CM-  CM=  CM\ 


EXERCISES 


1.  Show  that  two  points  of  a  straight  line  which  are  harmonically 
separated  by  two  others  are  both  on  the  same  side  of  the  mid-point  of  the 
line-segment  determined  by  the  two  others. 

2.  The  base  of  any  triangle  is  cut  harmonically  by  the  bisectors  of  the 
internal  and  external  vertical  angles. 

3.  The  hypotenuse  of  a  right  triangle  is  cut  harmonically  by  two  lines 
through  the  vertex  of  the  right  angle  which  make  equal  angles  with  one  of 
the  sides. 

^  4.  A  straight  line  meets  two  intersecting  circles  at  P,  Q,  R,  S,  and 
their  common  chord  at  0;  prove  that  OP,  OQ,  OB,  OS,  taken  in  a 
proper  order,  are  proportional. 


SIMILAR  RECTILINEAR  FIGURES  189 


MISCELLANEOUS   EXERCISES 

1.  If  two  triangles  have  one  angle  of  the  one  equal  to  one  angle  of  the 
other,  and  a  second  angle  of  the  one  supplementary  to  a  second  angle  of 
the  other,  then  the  sides  about  the  third  angles  are  proportional. 

2.  AE  bisects  the  vertical  angle  of  the  triangle  ABC  and  meets  the 
base  in  E ;  show  that  if  circles  are  described  about  the  triangles  ABE 
and  ACE,  their  diameters  are  to  each  other  in  the  same  ratio  as  the 
segments  of  the  base. 

3.  Two  circles  touch  internally  at  O  ;  AB  a  chord  of  the  larger  circle 
touches  the  smaller  at  C  which  is  cut  by  the  lines  OA,  OB  at  P  and  Q  ; 
show  that  OP:  OQ  =  AC:CB. 

4.  If  two  triangles  have  their  sides  parallel  in  pairs,  the  straight  lines 
joining  their  vertices  meet  in  a  point,  or  are  parallel. 

5.  If  any  two  similar  polygons  have  three  pairs  of  corresponding  sides 
parallel,  the  straight  lines  joining  the  corresponding  vertices  meet  in  a 
point  or  are  parallel. 

6.  If  A,  B,  C,  D  are  any  four  points  on  a  circle  and  E,  F,  G,  H  are 
the  mid-points  of  the  arcs  AB,  BC,  CD,  DA,  respectively,  prove  that  the 
straight  lines  EG  and  FH  are  at  right  angles. 

7.  The  sum  of  the  perpendiculars  drawn  from  any  point  within  an 
equilateral  triangle  on  the  three  sides  is  invariable. 

8.  Prove  that  the  straight  lines  which  trisect  one  angle  of  a  triangle 
do  not  trisect  the  opposite  side. 

9.  That  part  of  any  tangent  to  a  circle  which  is  intercepted  between 
tangents  at  the  extremities  of  a  diameter  is  divided  at  the  point  of  con- 
tact into  segments  such  that  the  radius  of  the  circle  is  a  mean  proportional 
between  them. 

10.  If  two  chords  AB  and  A  C,  drawn  from  a  point  ^  on  a  circle  ABC, 
are  produced  to  meet  the  tangent  at  the  other  extremity  of  the  diameter 
through  A,  in  the  points  D  and  E  respectively,  show  that  the  triangle 
AED  is  similar  to  the  triangle  ABC. 

11.  On  a  circle  of  which  AB  is  a  diameter  take  any  point  P.  Draw 
PC  and  PD  on  opposite  sides  of  AP  and  equally  inclined  to  it,  meeting 
AB  at  C  and  D.    Prove  AC:BC  =  AD:  BD. 


190  ELEMENTARY  GEOMETRY  [Chap.  Ill 

SUMMARY   OF   CHAPTER  III 
1.    Definitions. 

(1)  To  Measure  —  to  find  out  by  experiment  how  many  times  a 

given  magnitude  will  contain  a  chosen  unit.     §  211. 

(2)  Multiple  of  a  Given  Magnitude  —  a  magnitude  which  will  con- 

tain that  magnitude  an  integral  number  of  times.     §  211. 

(3)  Measure  of  a   Given  Magnitude  —  a  magnitude   which  is  con- 

tained in  that  magnitude  an  integral  number  of  times.     §  211. 

(4)  Commensurable  Magnitudes  —  such  as  can  be  measured  with  a 

common  unit.     §  213. 

(5)  Incommensurable  Magnitudes  —  such  as  cannot  be  measured  by 

any  common  unit.     §  213. 

(6)  Ratio  of  Two  Quantities  —  their  relative   magnitude,  i.e.  how 

many  times  one  is  as  great  as  the  other.     §  216. 

(7)  Ratio  of  Two  Commensurable  Magnitudes  —  the  ratio  of  their 

numerical  measures  by  a  common  unit.     §  220. 

(8)  Ratio  of  Two  Incommensurable  Magnitudes  —  the  limit  which 

the  ratio  of  their  approximate  measures  by  a  common  unit 
approaches,  as  this  unit  is  indefinitely  diminished.     §  229. 

(9)  Limit  of  a   Variable   Quantity  —  a  fixed  quantity  to  which  the 

variable  approaches  nearer  than  for  any  assignable  difference, 
though  it  cannot  be  made  absolutely  identical  with  it,     §  227. 

(10)  Proportion  —  a  statement  of  the  equality  of  two  ratios.     §  232. 

(11)  Continued  proportion.,  mean  proportional^   third  proportional. 

See  §  236. 

(12)  Mutually  Equiangular  Polygons — those  having    their   angles 

equal,  each  to  each,  and  in  the  same  order.     §  247. 

(13)  Similar  Polygons  —  two  polygons  which  have  the  same  number 

of  sides,  are  mutually  equiangular,  and  have  their  pairs  of  cor- 
responding sides  proportional.     §  248. 

(14)  Radical  Axis  of  Two  Intersecting   Circles  —  the  line  of  their 

common  chord.      §  263. 

(15)  Coaxial  System  of  Circles  —  the  circles  through  two  fixed  points. 

§  2(53. 

(16)  Division  in  Extreme  and  Mean  Ratio  — division  of  a  line-segment 

into  two  parts,  such  that  one  of  them  is  a  mean  proportional 
between  the  whole  segment  and  the  other  part.     §  274. 

(17)  Harmonic  Division  —  division  of  a  line-segment  internally  and 

externally  in  the  same  ratio.     §  283. 


Summary]         SIMILAR  RECTILINEAR  FIGURES  191 

2.  Postulates. 

(1)  If  P  and  Q  are  any  two  equal  magnitudes,  and  i?  is  a  third  mag- 

nitude of  the  same  kind,  then  the  ratio  oi  Pto  R  is  equal  to  the 
ratio  of  Q  to  i?,  i.e.  if  F=  Q,  then  P:R=  Q:R;  and  con- 
versely, if  P  and  Q  are  such  that  F:R=  Q:R^  then  F=  Q. 
(Postulate  6.)     §  218. 

(2)  If  P  and  Q  are  two  unequal  magnitudes,  and  P  is  a  third  magni- 

tude of  the  same  kind,  then  the  ratio  of  P  to  P  is  greater  or 
less  than  the  ratio  of  Q  to  P,  according  as  P  is  greater  or  less 
than  Q.     (Postulate  7.)     §  218. 

3.  Problems. 

(1)  To  divide  a  given  line-segment  into  any  required  number  of  equal 

parts.     §  269. 

(2)  To  divide  a  given  line-segment  into  parts  proportional  to  other 

given  line-segments.     §  270. 

(3)  To  divide  a  given  line-segment  internally   or  externally   in   a 

given  ratio.     §  271. 

(4)  To  find   a  fourth  proportional   to   three   given    line-segments. 

§  272. 

(5)  To  find  a  mean  proportional  between  two  given  line-segments. 

§273. 

(6)  To  divide  a  given  line-segment  in  extreme  and  mean  ratio.     §  275. 

(7)  Upon  a  given  line-segment  to  construct  a  polygon  similar  to  a 

given  polygon,  and  such  that  the  given  line-segment  shall  be 
homologous  to  a  given  side.     §  276. 

(8)  To  find  the  locus  of  a  point  whose  distances  from  two  fixed  points 

are  in  a  constant  ratio  different  from  unity.     §  282. 

4.  Theorem  on  Limits. 

If  there  are  two  variable  quantities  dependent  on  the  same  quantity 
in  such  a  way  that  they  remain  always  equal  while  each 
approaches  a  limit,  then  their  limits  are  equal.     §  230. 

5.  Theorems  on  Proportion. 

(1)  If  four  numbers  are  in  proportion,  the  product  of  the  extremes 

equals  the  product  of  the  means.     §  233, 

(2)  If  a:b  =  c:d,  then  by  inversion  b  :a  =  d:c,  and  by  alternation 

a:C  =  b:d.     §  234. 

(3)  It  a:b  =  c:d,  then  by  composition  a  +  b  :b  =  c  -\-  did,  by  divi- 

sion a  —  b  :b  =  c  —  d:d,  and  by  composition  and  division 
a-\-b:a-b  =  c  +  d:c-d.     §  237. 


192  ELEMENTARY  GEOMETRY  [Chap.  Ill 

(4)  If  three  terms  of  one  proportion  are  equal,  respectively,  to  the 

three  corresponding  terms  of  another,  their  fourth  terms  must 
be  equal.     §  238. 

(5)  If  any  number  of  ratios  are  equal,  then  the  sum  of  all  the  ante- 

cedents is  to  the  sum  of  all  the  consequents  njs  any  one  antece- 
dent is  to  its  consequent.     §  240. 

6.  Theorems  on  the  Similarity  of  Triangles. 

(1)  Two  triangles  are  similar  if  the  three  angles  of  one  are  equal, 

respectively,  to  the  three  angles  of  the  other.     §  249. 

(2)  If  two  triangles  have  two  angles  of  one  equal,  respectively,  to  two 

angles  of  the  other,  the  triangles  are  similar.     §  250. 

(3)  If  two  triangles  have  an  angle  of  one  equal  to  an  angle  of  the 

other,  and  the  sides  including  these  angles  proportional,  the 
triangles  are  similar.     §  251. 

(4)  If  the  ratios  of  the  three  sides  of  one  triangle  to  the  three  sides 

of  another,  two  and  two,  are  equal,  the  triangles  are  similar. 
§253. 

7.  Theorems  on  Chords  of  Circles. 

(1)  If  two  chords  of  a  circle  intersect  within  the  circle,  the  product 

of  the  segments  of  the  one  equals  the  product  of  the  segments 
of  the  other.     §  258. 

(2)  If  through  a  fixed  point  within  a  circle  any  chord  is  drawn,  the 

product  of  its  segments  is  the  same  whatever  its  direction. 
§259. 

(3)  Either  segment  of  the  least  chord  that  can  be  drawn  through  a 

fixed  point  within  a  circle  is  a  mean  proportional  between  the 
segments  of  any  other  chord  drawn  through  that  point.    §  260. 

(4)  A  perpendicular  drawn  from  any  point  of  a  circle  to  a  diameter 

is  a  mean  proportional  between  the  segments  into  which  it 
divides  the  diameter.     §  256. 

(5)  If  two  chords  of  a  circle,  when  produced,  intersect  without  the 

circle,  the  product  of  the  segments  of  the   one   equals  the 
product  of  the  segments  of  the  other.      §  261. 

(6)  If  a  tangent  and  a  secant  of  a  circle  intersect,  the  tangent  is  a 

mean  proportional  between  the  whole  secant  and  its  external 
segment.     §  262, 

(7)  If  from  any  point  on  the  common  chord  of  two  intersecting 

circles  produced,  tangents  are  drawn  to  the  circles,  the  lengths 
of  these  tangents  are  equal.     §  263. 


Summary]         SIMILAR   RECTILINEAR  FIGURES  193 

8.    Miscellaneous  Theorems. 

(1)  A  straight  line  parallel  to  one  side  of  a  triangle  divides  the  other 

two  sides  in  the  same  ratio,  and  conversely.     §§  242,  244. 

(2)  The  bisector  of  an  angle  of  a  triangle  divides  the  opposite  side 

into  segments  proportional  to  the  adjacent  sides.     §  245. 

(3)  If  two  similar  polygons  are  divided  into  triangles  by  the  diagonals 

drawn  from  a  pair  of  homologous  vertices,  these  triangles  are 
similar  each  to  each  and  are  similarly  placed,  and  conversely. 
§  252. 

(4)  The  perimeters  of  two  similar  polygons  are  in  the  same  ratio  as 

any  two  corresponding  sides.     §  254. 

(5)  If  in  a  right  triangle  the  perpendicular  is  drawn  from  the  vertex 

of  the  right  angle  to  the  hypotenuse :  (1)  the  two  triangles 
thus  formed  are  similar  to  each  other  and  to  the  whole  tri- 
angle ;  (2)  the  perpendicular  is  a  mean  proportional  between 
the  segments  of  the  hypotenuse  ;  (3)  each  side  of  the  triangle 
is  a  mean  proportional  between  the  hypotenuse  and  the  segment 
adjacent  to  that  side.     §  255. 

(6)  In  equal  circles,  angles  at  the  centre  are  in  the  same  ratio  as  the 

arcs  subtending  them,     §  266. 

(7)  If  the  sides  of  a  triangle  are  cut  by  any  straight  line,  the  product 

of  the  ratios  of  their  segments  taken  in  order  equals  unity,  and 
conversely  (Menelaus's  theorem).     §§  277,  278. 

(8)  If  two  triangles  are  so  situated  that  the  straight  lines  joining  the 

pairs  of  corresponding  vertices  are  concurrent,  then  the  points 
of  intersection  of  the  three  pairs  of  corresponding  sides  are 
collinear,  and  conversely  (Desargues's  theorem).     §§  279,  280. 

(9)  If  through  any  point  straight  lines  are  drawn  to  the  vertices  of  a 

triangle,  intersecting  the  sides,  the  product  of  the  ratios  of  the 
segments  of  the  sides  taken  in  order  equals  unity.     §  281. 
(10)  If  any  line-segment  AB  is  divided  harmonically  at  C  and  D,  the 
line-segment  CD  is  also  divided  harmonically  at  A  and  B.   §  284. 


CHAPTER    IV 
AREAS  OF  PLANE   POLYGONS 

Section  I 

PARALLELOGRAMS  AND  TRIANGLES 

287.  In  the  preceding  chapters  we  have  discussed  for  the 
most  part  only  geometrical  figures  (combinations  of  points  and 
lines)  and  the  conditions  under  which  two  such  figures  can  be 
made  to  coincide.  We  have  spoken  of  the  length  of  a  line- 
segment  and  of  the  relative  lengths  of  two  line-segments,  of  the 
size  of  an  angle  and  of  the  relative  sizes  of  two  angles ;  but  so 
far  have  had  nothing  to  do  with  surfaces  or  the  measurement 
of  surfaces. 

In  this  chapter  we  shall  consider  the  measurement  of  sur- 
faces enclosed  by  certain  plane  rectilinear  figures. 

288.  When  two  polygons  are  placed  so  as  to  have  one  or 
more  sides  or  parts  of  sides 
in  common,  without  over- 
lapping, they  are  said  to 
be  adjacent. 

By  disregarding  the  com- 
mon parts  of  the  boundaries 
of  two  adjacent  polygons,  a  third  polygon  is  obtained  which 
is  defined  to  be  the  sum  of  the  first  two. 

The  sum  of  two  polygons  is  thus  a  polygon  such  that  within 

194 


287-291]  AREAS  OF  PLANE  POLYGONS  195 

it  lie  all  the  points  which  are  inside  of  either  of  the  first  two, 
and  no  other  points. 

If  we  have  two  polygons  A  and  B  whose  sum  as  defined 
above  is  C,  then  the  difference  between  the  polygons  C  and  A 
is  the  polygon  By  or  the  difference  between  the  polygons  C  and 
B  is  the  polygon  A, 

289.  Definition.  The  area  of  a  plane  closed  figure  is  the 
surface  lying  within  the  figure. 

The  area  of  a  circle  is  the  surface  lying  within  the  circle,  i.e.  enclosed 
by  the  circle. 

The  area  of  a  triangle  is  the  surface  enclosed  by  the  three  sides  of  the 
triangle. 

Axiom  11.  If  tivo  plane  polygons  or  other  closed  figures  are 
identically  equal,  their  areas  are  equal. 

Axiom  12.  The  sum  of  the  areas  of  two  plane  polygons  is 
equal  to  the  area  of  their  sum. 

290.  To  measure  a  surface,  we  must  first  fix  upon  a  unit  of 
surface. 

The  unit  most  frequently  chosen  is  the  surface  enclosed  by 
a  square  whose  sides  are  each  of  unit  length. 

If  the  unit  length  is  one  inch,  the  unit  of  surface  is  the  surface  enclosed 
by  a  square  each  of  whose  sides  is  an  inch  in  length.  Such  a  surface  is 
called  a  square  inch. 

If  the  unit  of  length  is  a  foot,  then  the  unit  of  surface  is  a  square  foot ; 
or,  if  the  unit  of  length  is  a  yard,  or  a  centimetre,  the  unit  of  surface  is  a 
square  yard,  or  a  square  centimetre,  and  so  on. 

291.  The  measure  of  the  area  of  any  geometrical  figure  is 
the  number  which  expresses  how  many  times  this  area  will 
contain  the  chosen  unit  of  surface. 

For  brevity  we  shall  frequently  speak  of  '  the  area  of  a  figure'  as  though 
it  were  'the  measure  of  the  area,'  and  in  that  sense  the  area  is  a  number. 

Notation.  Hereafter  the  symbol  ZZ7  will  frequently  be  used  as  an 
abbreviation  for  the  word  '  parallelogram,'  |  |  for  the  word  'rectangle,' 
and  □  for  the  word  '  square.' 


196  ELEMENTARY  GEOMETRY  [Chap.  IV 


Proposition  I 

292.  Paralleloirams  upon  the  same  base  and  between 
the  same  parallels  are  equal  in  area. 


Let  ABCD  and  ABC'D'  be  two  parallelograms  on  the  same 
base  AB,  such  that  the  sides  opposite  to  AB,  viz.  CD  and  C'B'j 
are  in  the  same  straight  line. 

It  is  required  to  prove  that  these  parallelograms  are  equal  in 
area. 

Proof.  The  whole  polygon  ABC'D  is  the  sum  of  CJ  ABCD 
and  A  BC'C  (Def.) 

It  is  also  the  sum  of  O  ABC'D'  and  A  AD'D.  (Def.) 

Therefore  the  area  of  O  ABCD  +  the  area  of  A  BC'C  =  the 
area  of  O  ABC'D'  +  the  area  of  A  AD'D.  (Ax.  12.) 

But  As  BC'C  and  AD'D  are  identically  equal.     Prove. 

Therefore  the  area  of  A  BC'C  =  the  area  of  A  AD'D. 

(Ax.  11.) 

Hence  the  area  of  O  ABCD  =  the  area  of  O  ABC'D'. 

Notice  that  the  proof  applies  equally  well  to  any  of  the  three  possible 
cases  represented  in  the  diagrams,  viz.  (1)  when  DC  and  D'C  overlap, 
(2)  when  C  and  D'  coincide,  (3)  when  DC  and  D'C  are  separated. 

293.  When  we  say  that  two  polygons  are  equal,  we  shall 
mean  simply  that  they  have  equal  areas,  and  not  that  they  are 
necessarily  superposable.  If  it  is  meant  that  two  figures 
are  superposable,  we  shall  say  as  heretofore  that  they  are  iden- 
tically equal. 

294.  Corollary  I.  Parallelograms  upon  eqiial  bases  and 
between  the  same  parallels  are  equal  in  area. 


292-298]  AREAS   OF  PLANE  POLYGONS  197 

Let  ABCD  and  A'B'C'D'  be  two  parallelograms  upon  equal  bases  and 
between  the  same  parallels.  Join  A'D  and  B'G.  A'B'CD  is  also  a 
parallelogram.     Why  ? 

V  C  D'  & 

Then  UABCD  =0  A'B'CD.  /"--^^    /^--^^^     \  V 

For  they  are  upon  the  same  base      /             /        ^"^"^^^    \^^^->^  \ 
CD  and  between  the  same  parallels.    J ^ ^^^^jp ^^/ 

Similarly  O  A'B'  CD  =  O  A'B'  CD'. 

Therefore  O  ABCD  =  OA'B'  CD'. 

295.  Corollary  II.  If  a  triangle  arid  a  parallelogram,  are 
ujjon  the  same  base,  or  upon  equal  bases,  and  between  the  same 
parallels,  the  area  of  the  triangle  equals  half  the  area  of  the 
parallelogram. 

Let  A  ABE  and  O  ABCD  be  upon  the  same,  or  equal  bases,  and 
between  the  same  parallels.  n      w  w 

It  is  required  to  prove  that  the  area  of  the 
triangle  equals  half  the  area  of  the  parallelo- 
gram.    Complete  O^BF^. 


Then  CJ  ABCD  =  O  ABFE.     (Prop.  I. )     ^ 


Also  A  ABE  is  identically  equal  to  A  FEB  (Art.  124)  and  its  area  is 
therefore  half  the  area  of  CJ  ABFE. 

Therefore  the  area  of  A  ABE  equals  half  the  area  of  O  ABCD. 

296.  Corollary  III.  Triangles  upon  the  same  base,  or  upon 
equal  bases,  and  between  the  same  parallels  are  equal  in  area. 

297.  Corollary  IV.  Triangles  upon  equal  bases  in  the 
same  straight  line,  having  their  opposite  vertices  in  coynmon,  are 
equal  in  area. 

Definitions 

298.  When  one  side  of  a  parallelogram  has  been  named  the 
base,  the  perpendicular  distance  between  it  and  the  opposite 
side  is  called  the  altitude  of  the  parallelogram. 


198  ELEMENT ABY  GEOMETRY  [Chap.  IV 

When  one  side  of  a  triangle  has  been  named  the  base, 
the  perpendicular  distance  of  the  opposite  vertex  from  it  is 
called  the  altitude  of  the  triangle. 

299.  Corollary  V.  Any  parallelogram  is  equal  in  area  to 
a  rectangle  having  an  equal  base  and  an  equal  altitude. 


EXERCISES 

1.  The  straight  line  joining  any  vertex  of  a  triangle  to  the  mid-point 
of  the  opposite  side  forms  with  the  sides  two  triangles  of  equal  areas. 

2.  Of  two  triangles  which  are  between  the  same  parallels,  that  has  the 
greater  area  which  has  the  greater  base. 

3.  If  two  triangles,  equal  in  area,  are  upon  the  same  base  and  upon 
the  same  side  of  it,  the  straight  line  joining  their  vertices  is  parallel  to  the 
base. 

4.  ABC  is  any  triangle  and  DE  is  drawn  parallel  to  the  base  BC, 
meeting  AB  and  AC  ?it  D  and  E ;  BE  and  CD  aire  joined.  Prove  that 
A  DBC  =  A EBC,  A  BDE  =  A  CED,  and  A  ABE  =  AACD. 

5.  ABCD  is  a  quadrilateral  having  AB  parallel  to  CD;  its  diagonals 
AC  and  BD  intersect  at  O.     Prove  that  AAOD  =  ABOG. 

6.  If  through  the  vertices  of  a  triangle,  straight  lines  are  drawn  par- 
allel to  the  opposite  sides  and  produced  to  meet,  the  triangle  so  formed  is 
the  sum  of  four  equal  triangles. 

7.  On  the  same  base  and  between  the  same  parallels  as  a  given  paral- 
lelogram, construct  a  rhombus  equal  in  area  to  the  given  parallelogram. 

8.  Divide  a  given  parallelogram  into  two  equal  parallelograms.  In 
how  many  ways  can  this  be  done  ? 

9.  ABC  is  any  triangle,  and  D  any  point  in  AB.  Find  a  point  E  in 
BC  produced  such  that  ADBE=AABC. 

10.  If  one  diagonal  of  a  quadrilateral  bisects  the  other  diagonal,  it 
divides  the  quadrilateral  into  two  triangles  equal  in  area. 

11.  ABCD  is  any  quadrilateral,  AC  and  BD  its  diagonals.  A  paral- 
lelogram EFGH  is  constructed  by  drawing  through  the  vertices  straight 
lines  parallel  to  the  diagonals.    Prove  that  ABCD  equals  half  of  EFGH. 


298-300] 


AREAS   OF  PLANE  POLYGONS 


199 


Proposition  II 

300.   The  areas  of  two  rectangles  having  equal  alti- 
tudes are  in  the  same  ratio  as  their  bases. 


D 


G 


E 


Let  ABCD  and  EFGH  be  two  rectangles  of  equal  altitudes 
AD  and  EH. 


It  is  required  to  prove  that 

area  of  ABCD  :  area  of  EFGH 


AB :  EF. 


Proof.     There  are  two  cases  to  be  considered  according  as 
AB  and  EF  are  commensurable  or  incommensurable. 

Case  I.     When  AB  and  EF  are  commensurable. 
Let  AB  and  EF  have  a  common  measure  x  which  is  contained 
m  times  in  AB  and  n  times  in  EF. 


Then 


AB 
EF 


(Art.  220.) 


Divide  AB  and  EF  into  segments  each  equal  to  x,  and 
through  the  points  of  division  draw  lines  parallel  to  AD  and 
EH,  respectively,  thus  dividing  the  rectangle  ABCD  into  m 
rectangles,  and  the  rectangle  EFGH  into  n  rectangles,  all  identi- 
cally equal  (Art.  129),  and  therefore  equal  in  area. 

Since  I  I  ABCD  is  the  sum  of  m  of  these  equal  rectangles, 
and  I      I  EFGH  is  the  sum  of  n  of  them, 

area  of  ABCD  _  m 
area  of  EFGH  ~  n 

area  of  ABCD  ^  AB 
area  of  EFGH  ~  EF 

or,       area  of  ABCD  :  area  of  EFGH  =  AB :  EF. 


Therefore 


200  ELEMENTARY  GEOMETRY  [Chap.  IV 

Case  II.     When  AB  and  EF  are  incommensurable. 

Take  any  measure  of  EF,  say  x,  and  apply  it  to  AB  as  often 
as  possible,  letting  B'  be  the  last  point  of  division,  so  that  B'B 
is  a  remainder  less  than  x. 

J)  CfO       H  G 


A  B'B      E  .F 

Through  B'  draw  B'C  parallel  to  BC. 

Then  AB'C'D  and  EFGH  are  two  rectangles  of  equal  alti- 
tudes whose  bases  AB'  and  EF  are  commensurable. 

rru      f  area  of  AB'C'D     AB'  ,p^^^  t  ^ 

Therefore       = (Case  1.) 

area  of  EFGH      EF  ^  ^ 

If  now  the  measure  x  be  repeatedly  diminished,  the  base 
AB'  approaches  the  base  AB  as  its  limit,  while  the  rectangle 
AB'C'D  approaches  the  rectangle  ABCD  as  its  limit. 

Therefore  the  limit  of  the  variable  ratio  — .    ^^^.^ 

area  of  EFGH 

is  the  ratio  area  of  ABCD    ^^^  ^^^  j.^^.^  ^^  ^-^^  variable  ratio 

area  of  EFGH 

AB'  •    ,.         ,.     AB 

IS  the  ratio  — — -• 

EF  EF 

But  these  variable  ratios  are  always  equal ;  therefore  their 

limits  are  equal.     That  is 

area  of  ABCD  ^  AB  ,^^,^  230.) 

avesioiEFGH      EF  ^  ^ 

Or,  area  of  ABCD  :  area  of  EFGH  =  AB :  EF. 

301.  Corollary  I.  The  areas  of  two  rectangles  having  equal 
bases  are  in  the  same  ratio  as  their  altitudes. 

302.  Corollary  II.  Tlie  areas  of  two  triangles  having  equal 
altitudes  are  in  the  same  ratio  as  their  bases  ;  or,  having  equal 
bases  are  in  the  same  ratio  as  their  altitudes. 


300-303J 


AREAS   OF  PLANE  POLYGONS 


201 


Proposition  III 

303.   TJie  areas  of  any  two  rectangles  are  in  the  same 
ratio  as  the  products  of  their  bases  and  altitudes. 


D 


d 


A 


IT 


Let  ABCD  and  A'B'C'D'  be  two  rectangles  whose  areas  are 
denoted  by  P  and  P',  whose  bases  are  b  and  b',  and  whose 
altitudes  are  h  and  h',  respectively. 

It  is  required  to  prove  that   —  =  — — -• 

P'     bh' 

Proof.  Construct  a  third  rectangle,  denoting  its  area  by  Q, 
whose  base  is  equal  to  b',  and  altitude  equal  to  h. 

h'' 
P  ^  bh 
P'     b'h'' 

or  P:P'  =  bh:b'h'. 


Then 
Therefore 


P       ^  A 


(Prop.  II.) 
(Art.  239.) 


EXERCISES 

1.  One  rectangle  has  a  base  of  12  ft.  and  an  altitude  of  3  ft,,  another  a 
base  of  7  ft.  and  an  altitude  of  5  ft.     What  is  the  ratio  of  their  areas  ? 

2.  The  ratio  of  the  area  of  one  rectangle  to  the  area  of  another  is  3  :  2. 
The  first  has  an  altitude  of  8  ft.  and  a  base  of  15  ft.  The  second  has  an 
altitude  of  5  ft.     Find  its  base. 

3.  In  two  rectangles,  the  base  of  the  first  is  double  that  of  the  second, 
but  the  altitude  of  the  first  is  only  one-third  that  of  the  second  ;  what  is 
the  ratio  of  their  areas  ? 

4.  The  lengths  of  two  rectangular  fields  are  in  the  ratio  2  : 3.  What 
must  be  the  ratio  of  their  widths  in  order  that  the  first  may  have  an 
area  twice  as  great  as  the  second  ? 


202  ELEMENTABY  GEOMETRY  [Chap.  IV 

304.  Suppose  that  in  the  last  proposition  the  rectangle 
A'B'C'D'  is  a  square  having  unit  base  and  unit  altitude. 

Then  P'  is  the  unit  of  surface;  in  other  words,  it  is  unit 
area.      And  since  b'  and  h'  are  in  that  case  each  unity,  the 

relation  P:P'  =  bh:b'h' 

becomes  P :  unit  area  =  6/i :  1, 

or  P=bh  times  unit  area. 

That  is,  the   area  of  the  rectangle  ABCD  is  equal  to  the 
product  of  its  base  and  altitude  times  unit  area. 

In  other  words,  P  contains  the  unit  area  bh  times,  or  the 
measure  of  P  is  bh. 

Theorem.  Tlie  measure  of  the  area  of  a  rectangle  is  equal 
to  the  product  of  its  base  and  its  altitude. 

Or,  more  briefly,  the  area  of  a  rectangle  is  equal  to  the  product 
of  its  base  and  altitude. 

Exercise.  Divide  a  given  rectangle  into  unit  squares  by  straight 
lines  drawn  through  points  of  division  in  its  sides,  and  show  that  it  con- 
tains as  many  squares  as  is  represented  by  the  product  of  its  base  and 
altitude. 

305.  Corollary  I.  The  area  of  a  parallelogram  is  equal  to 
the  product  of  its  base  and  altitude. 

For,  any  parallelogram  is  equal  in  area  to  a  rectangle  having  the  same 
base  and  an  equal  altitude.  (Art.  299. ) 

306.  Corollary  II.  The  area  of  a  triangle  is  equal  to  half 
the  product  of  its  base  and  altitude. 

For,  the  area  of  a  triangle  is  half  the  area  of  a  parallelogram  having  the 
same  base  and  an  equal  altitude.  (Art.  295. ) 

307.  Corollary  III.  The  area  of  a  square  is  equal  to  the 
square  of  any  one  of  its  sides. 


304-309]  AREAS   OF  PLANE  POLYGONS  203 


Proposition  IV 

308.  The  areas  of  two  triangles  having  an  angle  of  the 
one  equal  to  an  angle  of  the  other  are  in  the  same  ratio 
as  the  products  of  the  sides  containing  the  equal  angles. 


B  G 

Let  BAC  and  B'AC  be  two  triangles  having  the  angles  at  A 
equal. 

It  is  required  to  prove  that  area  of  BAC:  area  of  BAC  = 
BA'AG'.B'A^AC. 

Proof.  Place  the  tria.ngles  so  that  the  equal  angles  coincide 
as  in  the  figure,  and  join  B'C. 

area  oi  BAC _  BA 
2ivesioiBAC~  B'A' 


Then  :       .;^.^  =  -^>  (Art.  302.) 


and 


Siresi  oi  BAC ^  AC 
Sivesioi  B'AC     AC' 


Therefore  area,  of  BAC       BA  -  AC  (Art.  239.) 

area  of  5'^ C"     BA  -  AC'  ^  ^ 


or        area  of  BAC :  area  of  B'AC  =  BA-AC  :  B'A  •  AC. 

309.  Corollary.  The  areas  of  two  jjarallelograms  having 
an  angle  of  the  one  equal  to  an  angle  of  the  other  are  in  the  same 
ratio  as  the  products  of  the  sides  containing  the  equal  angles. 

EXERCISE 

1.  ABCD  is  any  parallelogram.  Through  E^  any  point  in  the  diagonal 
^C,  straight  lines  are  drawn  parallel  to  the  sides.  Show  that  the  area  of 
the  parallelogram  EB  equals  the  area  of  the  parallelogram  ED. 


204  ELEMENTARY  GEOMETRY  [Chap.  IV 

310.  It  should  be  remembered  that  by  the  product  of  two 
line-segments  we  mean  the  product  of  their  measures,  and  that 
by  the  square  of  a  line-segment  we  mean  the  square  of  its 
measure. 

Thus  the  product  of  two  given  line-segments  equals  the  area 
of  a  rectangle  whose  adjacent  sides  are  respectively  equal  to 
the  given  segments ;  and  the  square  of  a  given  line-segment 
equals  the  area  of  a  square  each  of  whose  sides  equals  the 
given  segment. 

311.  Theorem.  If  a  given  line-segment  is  divided  iyiternally 
into  any  tivo  parts,  the  square  on  the  whole  segment  is  equal  in 
area  to  the  sum  of  the  squares  on  the  two  parts  together  with  twice 
the  rectangle  contained  by  the  two  parts. 

If  we  let  a  and  b  be  the  measures  of  the 
parts,  then  (a  -f  b)  is  the  measure  of  the 
given  segment  and  the  theorem  may  be 
stated  algebraically  thus : 

(a-i-by  =  a'-^b'-{-2ab. 

The  accompanying  diagram  illustrates 
geometrically  the  truth  of  the  theorem. 

312.  Theorem.  If  a  given  line-segment  is  divided  externally 
into  any  tvjo  parts,  the  square  on  the  given  segment  is  equal  in 
area  to  the  sum  of  the  squares  on  the  two  parts  less  twice  the 
rectangle  contained  by  the  two  parts. 

As  before,  let  a  and  b  be  the  measures  of  the  parts ;  then 
(a  —  b)  is  the  measure  of  the  given  segment,  and  the  theorem 
may  be  stated  algebraically  thus  : 

(a  -  by-  =  a^  +  &2  _  2  ab. 

Make  a  diagram  illustrating  this  theorem.     If  G  divides  AB 
externally,  and  you  let  AC  =  a,  and  BG  =  6, 
then  AB  —  a  —  b. 

Construct  the  squares  on  a  and  a—b  producing  the  sides  of 
the  latter  to  meet  the  former. 


"T 

ab 

b' 

i 

r  4 

a« 

ab 

. — — -^M 

310-313]  AREAS  OF  PLANE  POLYGONS  205 

313.   The  relation  between  tlie  two  theorems  stated  above 
can  be  best  shown  by  placing  them  side  by  side  thus : 

Theorem  (1)     a o £ 


Iff  =  AG''  +CB'  +  2AC'  CB. 
Theorem  (2)     a b c 

AB'==AC'-\-Cff-2AC'CB. 

The  two  theorems  are  identical  if  we  attach  a  sign  to  the 
segments.  If  C  divides  the  segment  AB  internally,  the  parts 
AC  and  CB  are  of  the  same  sign,  while  if  C  divides  AB  ex- 
ternally the  parts  are  of  opposite  signs. 

Thus  the  product  AC  •  CB  is  in  the  first  case  positive,  and 
in  the  second  case  negative.  The  square  of  either  a  positive 
or  a  negative  segment  is  of  course  positive. 

With  this  interpretation  the  relation 

Aff  =  AC'-{-CB'-{-2AC'  CB 

is   true  no  matter  how  the  points  A,  B,  C,  are   placed  on   a 
straight  line. 

This  relation  furnishes  us  with  another  illustration  of  the 
Principle  of  Continuity. 

EXERCISES 

1.  Illustrate  the  following  relations  geometrically  and  state  in  words 
the  theorem  contained  in  each. 

(1)  a(6  +  c)=  db  -{•  ac. 

(2)  «.  =  4(1)1 

(3)  a^-h^:=(a  +  b)ia-b). 

(4)  (rt  +  by  +  (a  -  &)2  =  2  a2  4.  2  &2. 

(5)  (a  +  by  -(a -by  =  4  ab. 


206  ELEMENTARY  GEOMETRY  [Chap.  IV 


EXERCISES 

1.  The  rectangle  contained  by  two  line-segments  is  equal  to  twice  the 
rectangle  contained  by  one  of  them  and  half  of  the  other. 

2.  If  a  line-segment  is  divided  internally  into  any  three  parts,  the 
square  on  the  whole  segment  is  equal  to  the  sum  of  the  rectangles  con- 
tained by  the  whole  segment  and  its  three  parts. 

3.  Show  that  if  a  square  and  a  rectangle  have  equal  perimeters  the 
square  has  the  greater  area. 

4.  If  through  the  vertices  of  any  triangle  straight  lines  parallel  to  the 
sides  are  drawn  to  meet,  two  and  two,  the  resulting  triangle  will  contain 
three  equal  parallelograms. 

5.  Equal  parallelograms  on  opposite  sides  of  the  same  base  are  of 
equal  altitude. 

6.  If  two  equal  triangles  are  upon  the  same  base  and  on  opposite  sides 
of  it,  the  straight  line  joining  their  vertices  is  bisected  by  the  base. 

7.  If  a  quadrilateral  is  divided  into  equal  triangles  by  each  of  its 
diagonals,  it  is  a  parallelogram. 

8.  On  the  base  of  a  given  triangle  construct  another  triangle  of  equal 
area  having  its  vertex  on  a  given  straight  line.  In  what  case  is  this  im- 
possible ? 

9.  The  area  of  a  trapezoid  is  equal  to  the  product  of  its  altitude  and 
half  the  sum  of  its  parallel  sides. 

10.  Show  that  the  sum  of  the  squares  on  the  two  segments  of  a  given 
line-segment  is  the  least  possible  when  it  is  bisected. 

11.  The  sum  of  the  squares  on  two  internal  segments  of  a  given  line- 
segment  becomes  greater  as  the  point  of  section  approaches  one  extremity. 

12.  Inscribe  a  square  in  a  given  semicircle. 

13.  If  A,  C,  Z>,  B  are  four  points  on  a  straight  line  so  situated  that  D 
bisects  OjB,  prove  that  the  square  on  AC  is  less  than  the  sum  of  the 
squares  on  AD  and  DB  by  twice  the  rectangle  AD  •  DB. 

14.  If  BAC  is  any  acute  angle  and  BD,  CE,  are  drawn  perpendicular 
to  its  boundaries  AC,  AB,  respectively,  show  that  the  rectangle  whose 
sides  are  equal  to  AB  and  AE  is  equal  in  area  to  the  rectangle  whose 
sides  are  equal  to  ^C  and  AD. 

15.  If  ^J?C  be  a  right-angled  triangle  and  CD  be  drawn  perpendicular 
to  the  hypotenuse,  then  AD  :  DB  =  ACf  :  BG^. 


313-314]  AREAS   OF  PLANE  POLYGONS  207 

Section   II 

AREAS    OF    SIMILAR    POLYGONS 

Proposition  V 

314.   The  areas  of  similar  triangles  are  in  the  same 
ratio  as  the  squares  of  any  two  homologous  sides. 


B'    D' 


Let  ^5(7  and  A'B'C  be  two  similar  triangles,  of  which  AB 
and  A'B'  are  homologous  sides. 
It  is  required  to  prove  that 

area  of  ABC  :  area  of  A'B'C  =  AB'  :  A^^- 
Proof.    Let  AD  and  A'D'  be  the  altitudes  of  the  two  triangles. 

T^^^    area  of  .150  ^  i  BC  ^  AD  ^B^,A^,     ^j,^,^  30e.) 
area  of  ^'^'C      ^B'C-A'D'      B'C    A'D'      ^  ^ 

Now  As  ABC  and  A'B'C  are  similar ; 
therefore  11  =  if." 

And  since  As  ADB  and  A'D'B'  are  also  similar,    (Art.  249.) 
AD       AB 


Therefore 


A'D'     A'B' 
Sivesioi  ABC       AB      AB       AB" 


area  of  yl'^'C      A'B^    A'B'     Jjb'' 


EXERCISES 


1.    If  the  mid-points  of  two  sides  of  a  triangle  are  joined  by  a  straight 
line,  what  part  of  the  whole  triangle  is  the  smaller  one  so  formed  ? 


208 


ELEMENTARY   GEOMETRY 


[Chap.  IV 


Proposition  VI 

315.   The  areas  of  similar  polygons  are  in  the  same 
ratio  as  tlve  squares  of  any  two  homologous  sides. 

B 


ED  E'  D 

Let  ABCDEF  and  A'B'C'D'E'F'  be  any  two  similar  polygons 
of  which  AB  and  A'B'  are  homologous  sides. 
It  is  required  to  prove  that 

area  oiAB'-F:  area  of  A'B'  ...  i^'  =  ab" :  A^'\ 

Proof.     Divide  the  two  polygons  into  triangles  by  drawing 
the  diagonals  from  two  homologous  vertices,  A  and  A'.     These 


triangles  will  be  similar,  two  and  two. 

Let  the  areas  of  the  triangles  be  P,  Q,  E,  S 
R',  S',  .... 


Then 


p_ab'     q_ac'_ab' 


A^' 


R 
R' 


AD 


Q'     A'C'     A'B 
AC       AB 


f2' 


(Art.  252.) 
and  P,  Q', 

(Prop.  V.) 


2  rp^ 


etc. 


A'B' 


Whence 


Therefore 


That  is, 


A'B' 


A'D'^     A'C" 

^=R=^=^=      = 

P      Q'     R'     S'      '"      A^' 

P^Q^R^S  +  -"  ^A^ 
P+Q'^-R'+S'+''      aJB'' 

areaof  ^JB...F  ^  Al^ 
areaof  ^'5'...i^'      ^Tgf2* 


(Art.  240.) 


316.   Corollary.     The  areas  of  similar  polygons  are  in  the 
same  ratio  as  the  squares  of  any  two  homologous  diagonals. 


315-317] 


ABE  AS   OF  PLANE  POLYGONS 


209 


Proposition  VII 
317.   The  area  of  the  square  described  on  the  hypot- 
enuse of  a  right  triangle  is  equal  to  the  sum   of  the 
areas  of  the  squares  described  on  the  other  two  sides. 


Let  ABG  be  a  triangle,  right-angled  at  C,  and  let  AH,  BD, 
and  CG  be  squares  described  upon  the  three  sides. 

It  is  required  to  prove  that  the  area  of  the  square  AH  is  equal 
to  the  sum  of  the  areas  of  the  two  squares  BD  and  CG. 

Proof.     Draw  CK  parallel  to  A  J  or  BH,  and  join  CJ  and  BG. 

FC  and  GB  are  in  the  same  straight  line.     Why  ? 

Therefore  A  GAB  =  half  of  D  GACF,  .     (Art.  295.) 

Also  A  AJC  =  half  of  IZZ]  AJKL. 

But  As  GAB  and  AJC  or  CAJ  are  identically  equal.    Why  ? 

Therefore  D  GACF  =  CD  AJKL. 

■  By  joining  AE  and  CH,  it  may  be  similarly  proved  that 
nBEDC  =  CI}BHKL. 

This  statement  .should  be  fully  verified  by  the  pupil. 

But  I      Is  AJKL  and  BHKL  together  make  up  D  AJHB. 
Therefore  the  areas  of  Ds  ACFG  and  BE  DC  are  together 
equal  to  the  area  of  n\AJHB. 

This  theorem   is   usually  attributed   to  Pythagoras,  a  Greek 
mathematician  who  lived  about  550  b.c. 


210 


ELEMENTARY  GEOMETRY 


[Chap.  IV 


Alternate  Proofs 

1.   In  the  triangle  ABC,  right-angled  at  C,  draw  CD  perpen- 
dicular to  the  hypotenuse. 


Then 


or 


Also 


(Art.  255.) 
(Art.  255.) 


AD  B 

AB:AC=AG:AD, 

ACf  =  AB'AD, 

AB:BC=BC:DB, 

or  BO^  =  AB'  DB. 

Therefore  AC^  -\- BC^  =  AB  -  AD  +  AB  -  DB 

=  AB(AD-{-DB) 

=  AB'AB 

=.AB\ 

2.   With  centre  A  and  radius  AC  describe  a  circle. 
Let  the  side  BA  meet  this  circle  in  the  points  E  and  F. 


Then  BC  is  tangent  to  the  circle.     Why  ? 
Hence  BE'BF=  BC\  (Art.  262.) 

But  BE  =  BA-EA=^BA-AC, 

and  BF===BA-\-AF=BA  +  AG. 


317-318]  AREAS  OF  PLANE  POLYGONS  211 

Therefore  BE  -  BF  =  {BA  -  AC)  {BA  +  AC)  =  BA'  -  AC'- 

Hence  BA^  -  AC^  ==  BC\ 

That  is,  BA'  =  AC'  +  BC\ 

Many  other  methods  for  proving  this  theorem  have  been  devised.  The 
proof  presented  in  the  main  proposition  is  that  found  in  Euclid's  Ele- 
ments. The  nature  of  the  proof  given  by  Pythagoras  is  not  certainly 
known. 

318.  Corollary.  The  area  of  any  polygon  described  on  the 
hypotenuse  of  a  right  triangle  is  equal  to  the  sum  of  the  areas  of 
the  similar  polygons  similarly  described  on  the  other  two  sides. 


Let  a,  b,  c,  be  the  sides  of  the  right  triangle,  and  A,  B,  C,  be 
the  areas  of  the  polygons  similarly  described  on  them. 

Then  A:C  =  a':c\  (Prop.  V.) 

B:C=b-':c\ 

Therefore  A-h B:  C  =  a' -{-b^ :  c".        ( Ex.  3,  p.  165.) 

But  a^  +  62  =  c^ ;  (Prop.  VII.) 

therefore  A  +  B=C. 

EXERCISES 

1.  State  and  prove  the  converse  of  Proposition  VII. 

2.  If  the  difference  of  the  squares  on  two  sides  of  a  triangle  is  equal 
to  the  square  on  the  third  side,  the  triangle  is  right-angled. 

3.  The  square  described  on  the  diagonal  of  a  given  square  is  equal  to 
twice  the  given  square. 


212  ELEMENTARY  GEOMETRY  [Chap.  IV 

4.  A  quadrilateral  is  such  that  its  diagonals  intersect  at  right  angles. 
Prove  that  the  sum  of  the  squares  on  one  pair  of  opposite  sides  is  equal  to 
the  sum  of  the  squares  on  the  other  pair  of  opposite  sides. 

6.  If  0  is  the  point  of  intersection  of  the  perpendiculars  drawn  from 
the  vertices  of  a  triangle  ABC  to  the  opposite  sides,  the  squares  on  OA 
and  BC  are  together  equal  to  the  squares  on  OB  and  CA,  and  also  to  the 
squares  on  0(7  and  AB. 

6.  If  the  hypotenuse  and  a  side  of  one  right  triangle  are  equal  to  the 
hypotenuse  and  a  side  of  another  right  triangle,  show  that  the  triangles 
are  identically  equal. 

Prove  by  use  of  Proposition  VII.     See  also  a  different  proof 
in  Chapter  I,  Proposition  XIX. 

7.  Find  a  line-segment  the  square  on  which  is  equal  to  three  times 
the  square  on  a  given  line-segment. 

8.  Find  a  line-segment  the  square  on  which  is  equal  to  the  difference 
of  the  squares  on  two  given  line-segments. 

9.  Five  times  the  square  on  the  hypotenuse  of  a  right  triangle  is  equal 
to  four  times  the  sum  of  the  squares  on  the  medians  to  the  other  two 
sides. 

10.  Three  times  the  square  on  any  side  of  an  equilateral  triangle  is 
equal  to  four  times  the  square  on  the  perpendicular  drawn  from  a  vertex 
to  the  opposite  side. 

PROJECTIONS 

319.  If  from  a  given  point  a  perpendicular  is  drawn  to  a 
given  straight  line,  the  point  of  intersection,  or,  as  we  say, 
the  foot  of  the  perpendicular,  is  called  the  projection  of  the  given 
point  upon  the  given  line. 

The  projection  of  the  point  F 
upon  the  straight  line  AB  is  the 
point  P',  and  the  projection  of 
Q  is  Q',  if  PF  and  QQ'  are  per- 


pendicular to  AB.  A  P'  Q'  B 

The  projection  of  a  given  line-segment  upon  any  straight  line 

is  the  intercept  between  the  projections  of  its  extremities  upon 
the  line. 


318-320] 


AREAS  OF  PLANE  POLYGONS 


213 


The  projection  of  the  line-segment  PQ  upon  the  straight  line 
AB  is  the  segment  P'Q';  the  projection  of  the  line-segment 
RS  upon  AB  is  the  segment  RS'. 

A  line-segment  is  equal  to  its  projection  upon  any  straight 
line  parallel  to  it,  but  is  greater  than  its  projection  upon  any 
strais^ht  line  not  narallel  to  it. 


straight  line  not  parallel  to  it. 


What  is  the  projection  upon  a  given  straight  line  of  a  segment  at  right 
angles  to  it  ? 

320.  Theorem.  If  two  line-segments  OA  and  OB  have  one 
extremity  in  common,  the  product  of  OA  and  the  projection  of 
OB  upon  the  line  of  OA  equals  the  product  of  OB  and  the 
projection  of  OA  upon  the  line  of  OB. 

From   A  and   B  draw  the   perpendiculars  AA^   and  BB\ 

respectively. 

A 

'A 


Then  OA'  is  the  projection  of  OA  upon  the  straight  line 
OB,  and  OB'  is  the  projection  of  OB  upon  the  straight  line 
OA,  and  we  are  required  to  prove  that  OA  •  OB'  =  OB  •  OA'. 

As  OAA'  and  OBB'  are  similar.     Whyf 


Therefore  OA  :  OA'  =  OB  :  OB', 

or  OA  '  OB'  =  OB  •  OA'. 


(See  Ex.  14,  p.  206.) 


214 


ELEMENTARY  GEOMETRY 


[Chap.  IV 


Proposition   VIII 

321.  In  any  triangle  the  square  on  the  side  opposite 
an  acute  angle  is  less  than  the  sum  of  the  squares  on 
the  other  two  sides  hy  twice  the  product  of  either  of 
these  sides  and  the  projection  of  the  other  upon  it. 


B'       A 

Let  the  angle  C  of  the  triangle  ABO  be  acute,  and  from  B 
draw  the  perpendicular  BB'  to  the  line  AC,  so  that  GB'  is 
the  projection  of  CB  upon  the  line  AC. 

It  is  required  to  prove  that 

AB"  =  AC'  +C^-2AC'  CB'. 

Proof.     Since  in  either  diagram  AB'  is  divided  externally 


at  C, 


AB''  =  AC'  +  CB''  -2  AC'  CB'.       (Art.  312.) 


To  each  side  of  this  relation  add  BB' . 

Then  AB''  +  BB''  =  AC'  +  CB''  +  BB^  -  2  AC  -  CB'. 

But  AB''  +  BB''=^  A&,  (Prop.  YII.) 

and  CB''  +  BB''  =  CB'.  (Prop.  VII.) 

Therefore     AS  =  AC'  -{- CB"  -  2  AC  -  CB'. 

What  does  this  relation  become  in  case  B'  coincides  with  A,  i.e.  in 
case  ^  is  a  right  angle  ? 

EXERCISE 

If  AB  and  CD  are  any  two  given  line-segments,  prove  that  the  product 
of  AB  and  the  projection  of  CD  upon  the  line  of  AB  equals  the  product  of 
CD  and  the  projection  of  AB  upon  the  line  of  CD. 


321^23]  AREAS  OF  PLANE  POLYGONS  215 


Proposition   IX 

322.  In  an  obtuse-angled  triangle  the  square  on  the 
side  opposite  the  obtuse  angle  is  greater  than  the  sum 
of  the  squares  on  the  other  two  sides  by  twice  the  product 
of  either  of  these  sides  and  the  projection  of  the  other 
upon  it' 


A  G         B 


Let  the  angle  C  of  the  triangle  ABC  be  obtuse,  and  from  B 
draw  the  perpendicular  BB'  to  the  line  AG,  so  that  CB'  is 
the  projection  of  CB  upon  the  line  AG. 

It  is  required  to  prove  that 

^^  =  JC'  +  CB'  +  2  AG'GB\ 
Proof.    Since  AB^  is  divided  internally  at  (7, 

JLB^  =  J^  +  W  +  2  AG'  GB'.        (Art.  311.) 
To  each  side  of  this  relation  add  BB'^. 
Then  J^'  +  BB'^  =  AG"  +  GB'""  +  BB'^  +  2  AG -OB'. 
But  AB'  +  BB^"  =  A^, 

and  CB^^  +  BB'^  =  GB". 

Therefore       AB"  =  AG' +  US' -\- 2  AG  -  GB'. 

323.  No  doubt  the  pupil  has  already  noticed  the  close 
similarity  between  the  proofs  of  Propositions  VIII  and  IX. 
In  fact,  except  for  the  change  of  sign  of  the  segment  GB',  the 
proofs  are  identical. 


216  ELEMENTARY  GEOMETRY  [Chap.  IV 

We  might  with  perfect  propriety  consider  the  theorem  of 
Proposition  IX  a  general  theorem  of  which  both  Propositions 
VII  and  YIII  are  special  cases,  though,  ^ 

of  course,  the  proof  of  Proposition  IX 
must  follow  that  of  Proposition  VII,  since  y/^ 

it  depends  upon  it.  /       I 

Suppose,  for  example,  in  the  diagram  y^         / 

for  Proposition  IX  we  rotate   the   side         ^/^  -^ 

CB  about  C  in  the  way  indicated  by  the      jL. L 

arrow-head.     The  angle  ACB  gradually  ^ 

decreases  in  magnitude,  while  the  point  B'  approaches  G  and 
coincides  with  C  when  Z  ACB  becomes  a  right  angle. 

In  this  position  the  segment  CB'  is  zero  and  the  product 
2  AC '  CB'  is  also  zero. 

The  relation   AB'  =  AG^  +  CB^ -h  2  AC  •  CB' 

thus  becomes      AB' =  AC  +  CR, 

which  is  the  relation  proved  for  a  right  triangle  in  Proposition 
VII. 

If  the  rotation  of  CB  about  C  continues,  Z  ACB  becomes 
acute,  B'  moves  in  between  C  and  A,  the  segment  CB'  changes 
sign,  and  we  have  the  relation 

A^  =  AG'-\-W-2AC'CB', 

which  was  proved  independently  in  Proposition  VIII. 

Here  again  is  an  illustration  of  the  Principle  of  Continuity. 


EXERCISE 

The  sum  of  the  squares  on  two  sides  of  any  triangle  is  equal  to  twice 
the  squares  on  half  the  third  side,  and  on  the  median  drawn  to  that 
side. 

Suggestion.  The  median  makes  in  general  one  acute  angle  and  one 
obtuse  angle  with  the  third  side.  Apply  Propositions  VIII  and  IX,  and 
add. 


323-324] 


AREAS   OF  PLANE  POLYGONS 


217 


The  Area  of  a  Triangle  in  Terms  of  its  Three  Sides 

324.  The  area  of  a  triangle  has  been  found  (Art.  306)  to  be 
half  the  product  of  its  base  and  altitude,  but  it  is  sometimes 
more  convenient  to  have  at  hand  a  formula  expressing  the  area 
in  terms  of  the  lengths  of  the  three  sides.  Such  a  formula  we 
are  now  ready  to  deduce. 


B  D  C 

Letter  the  triangle  ABC  symmetrically,  denoting  by  a  the 
length  of  the  side  opposite  the  vertex  A,  by  b  the  length  of 
the  side  opposite  the  vertex  B,  and  by  c  the  length  of  the 
side  opposite  the  vertex  C. 
•    Giving  proper  attention  to  the  sign  of  the  segment  CD, 

AB'  =  BC'-hGA'  +  2BC-  CD, 

(Prop.  VIII  and  IX.) 

or  c2  =  a2  +  &2  _^  2  a  .  CD. 


Whence, 


CD  = 


-¥ 


2a 


If  Z  O  is  acute,  the  segment  CD  is  negative  as  appears  from  the 
diagram. 

Now  BD=BC-\-  CD 

2a 

^  a2  ^  c^  _  52 

2a 
That  is  to  say,  whether  Z  (7  is  acute  or  obtuse,  the  projection 

of  the  side  c  upon  the  side  a  equals  — — — • 

2  a 


218  ELEMENTARY  GEOMETRY  [Chap.  IV 

Next,  ad'  =  aS-BT?  (Prop.  VII.) 

=  {AB  +  BD)  (AB  -  BD) 

^  (g  +  cf  -  &^  ^  b^-(a-cy 
2a  2  a 

^  (a  +  b-\-c)(a  —  b-^c)(a-\-b  —  c)(b  +  c-a) 
4  a' 

or  u4Z>  =  ^  V(aH-&  +  G)(a— 6+c)(a  +  &  — c)(6  +  c— ci).. 

This  expression  for  the  length  of  AD  can  be  written  more 
conveniently  if  we  let         ci  +  6  +  c  =  2  s, 

so  that  a-\-b  —  c  =  2s  —  2c  =  2{s  —  c), 

a-b-{-c  =  2s-2b  =  2(s-b), 

and  b-{-c  —  a  =  2s  —  2a  =  2{s  —  a). 

Substituting  these  values  in  the  expression  for  AD  gives 

AD=^V2s'2{s-a)-  2(s  -  b)  •  2{s  -  c) 
Z  a 


=  -  Vs(s  —  a)(s  —  b){s  —  c). 
a 


That  is,  the  length  of  the  perpendicular  upoi%  the  side  of  a  tri- 

angle  from  the  opposite  vertex  A  equals  -■\/s{s  —  d)(s  —  b){s  —  c). 

a 

Finally,  the  area  of  A  ABC  =  ^  BC  -  AD,  (Art.  306.) 

o 

=  1  a  •  -^s{s  —  a)(s  —  b)(s  —  c) 
a 


=  Vs(5  —  a)(s  —  b)(s  —  c). 


324]  AREAS   OF  PLANE  POLYGONS  219 


EXERCISES 

1.  What  does  the  formula  in  Article  324  for  the  projection  of  the 
side  c  upon  the  side  a  become  when  C  is  a  right  angle  ? 

2.  From  symmetry  write  out  the  projections  of  (1)  the  side  c  upon  the 
side  6,  (2)  the  side  a  upon  the  side  &,  etc. 

3.  By  symmetry,  write  out  the  expressions  for  the  lengths  of  the 
perpendiculars  from  the  vertices  B  and  C  upon  the  opposite  sides. 

4.  Show  that  the  shortest  perpendicular  falls  upon  the  longest  side  of 
the  triangle,  and  that  any  two  perpendiculars  are  inversely  proportional 
to  the  sides  upon  which  they  fall. 

5.  The  three  sides  of  a  triangle  are  58  ft.,  51  ft.,  and  41  ft.,  respec- 
tively, find  its  area. 

6.  Find  the  lengths  of  the  perpendiculars  from  the  three  vertices  on 
the  opposite  sides  of  a  triangle  whose  sides  are  7  in.,  9  in.,  and  11  in., 
respectively. 

7.  The  sides  a,  &,  c,  of  a  triangle  are  5,  9,  11  feet,  respectively ;  find 
whether  or  not  it  contains  an  obtuse  angle,  and  the  length  of  the  projec- 
tions of  the  sides  a  and  b  upon  the  side  c. 

8.  Find  the  area  in  acres  of  a  triangular  field  whose  sides  are  320, 
426,  and  261  yards,  respectively. 

9.  Find  the  sides  of  a  right  triangle  if  their  projections  upon  the 
hypotenuse  are  3if  and  13^7  feet,  respectively. 

10.  The  two  parallel  sides  of  a  trapezoid  are  16  and  21  feet,  respectively, 
and  their  distance  apart  is  5  feet.     Find  the  area. 

11.  The  sum  of  the  squares  on  the  four  sides  of  a  parallelogram  is  equal 
to  the  sum  of  the  squares  on  the  diagonals. 

12.  The  square  on  the  base  of  an  isosceles  triangle  is  equal  to  twice  the 
rectangle  contained  by  either  of  the  equal  sides  and  the  projection  on  it 
of  the  base. 

13.  If  the  square  on  one  side  of  a  triangle  is  less  than  the  sum  of 
the  squares  on  the  other  two  sides,  how  does  the  angle  contained  by  these 
two  sides  compare  with  a  right  angle  ? 

14.  The  base  of  a  triangle  being  given,  find  the  locus  of  the  vertex  when 
the  sum  of  the  squares  on  the  two  sides  is  also  given. 

Suggestion.     Apply  Ex.  1,  p.  216. 


220  ELEMENTARY  GEOMETRY  [Chap.  IV 

Section  III 

CONSTRUCTIONS 

Proposition   X 

325.  To  construct  a  rectangle  equal  in  area  to  a  given 
parallelogram,  and  having  one  side  equal  to  a  given 
line-segment. 

D'  D  &  O  S^ \(^ 


I^J 


A  B  E  F 

It  is  required  to  construct  a  rectangle  upon  the  given  line- 
segment  EF  which  shall  be  equal  in  area  to  the  given  paral- 
lelogram ABGD. 

Construction.  Draw  AD'  and  BC  perpendicular  to  CD,  thus 
forming  a  rectangle  ABC'D'  equal  in  area  to  the  given  paral- 
lelogram ABGD.  (Art.  299.) 

If  we  suppose  EFGH  to  be  the  required  rectangle,  of  which 
the  side  FG  is  as  yet  unknown, 

then  EF'FG  =  AB'BC\ 

since  the  area  of  a  rectangle  equals  the  product  of  two  adja- 
cent sides.  .  (Art.  304.) 
Therefore                 EF:AB  =  BC  :  FG 

and  the  required  side  FG  can  be  found  as  the  fourth  propor- 
tional to  EF,  AB,  and  BC.  (Art.  272.) 

The  length  of  the  side  FG  having  been  thus  found,  the  re- 
quired rectangle  is  easily  constructed  as  follows :  — 

At  E  and  F  draw  straight  lines  perpendicular  to  EF. 

Cut  off  FG  of  the  proper  length,  determined  above,  and  make 
EH  equal  to  it.     Join  HG. 

Prove  that  the  figure  as  constructed  is  a  rectangle. 


325-327]  AREAS   OF  PLANE  POLYGONS  221 

326.  The  method  of  analysis  employed  in  the  preceding  con- 
struction (see  page  86)  should  be  resorted  to  whenever  a  direct 
solution  of  a  problem  does  not  readily  appear. 

It  is  strongly  advised  that  the  pupil  actually  make  the  con- 
structions indicated  in  the  exercises,  drawing  all  lines  carefully 
with  ruler  and  compasses. 

EXERCISES 

1.  Construct  a  parallelogram  equal  in  area  to  a  given  triangle. 

2.  Construct  a  rectangle  equal  in  area  to  a  given  triangle  and  having 
one  side  equal  to  a  given  line-segment. 

3.  Construct  a  square  equal  in  area  to  a  given  parallelogram  or  triangle. 

Proposition   XI 

327.  To  construct  a  square  equal  in  area  to  the  sum 
of  two  given  squares. 


Let  a  and  h  be  sides  of  the  given  squares.  If  they  are 
placed  at  right  angles,  as  in  the  diagram,  then  c  will  equal  a 
side  of  the  required  square.     Why  ? 

EXERCISES 

1.  Construct  a  square  equal  in  area  to  the  difference  of  tw^o  given 
squares. 

2.  Construct  a  square  equal  in  area  to  the  sum  of  two  given  triangles. 

3.  Construct  a  square  which  shall  he  equal  to  the  sura  of  three  given 
squares. 

4.  Construct  a  rectangle  equal  in  area  to  a  given  square  and  having 
one  side  of  given  length. 


222 


ELEMENTARY  GEOMETRY 


[Chap.  IV 


Proposition   XII 

328.   To  find  two  line-segments  having  the  same  ratio 
as  the  areas  of  two  given  squares. 


Let  a  and  h  be  sides  of  the  given  squares. 

Place  these  lines  as  in  the  diagram  so  as  to  form  a  right 
angle  ACB  and  join  AB.  The  areas  of  the  given  squares  are 
in  the  ratio  BC  :  AC  . 


From  C  draw  CD  perpendicular  to  AB. 


Then 


(Art.  2^5.) 


or 


Also 


or 


Hence 


AD:AC=AC:AB, 

AG'  =  AD  .  AB. 

BD:BC=BC:BA, 

BC""  =  BD'BA  =  DB-  AB. 
BC' :  A~G'  =  DB'AB:AD'AB  =  DB:  AD. 


Therefore  DB  and  AD  are  the  required  line-segments. 


EXERCISES 

1.  Construct  a  square  which  shall  have  a  given  ratio  to  a  given  square. 

2.  If  c  is  the  third  proportional  to  a  and  &,  show  that  a  is  to  c  as  the 
square  on  a  is  to  the  square  on  b. 

3.  Find  two  line-segments,  one  of  given  length,  which  shall  be  in  the 
same  ratio  as  the  areas  of  two  given  squares. 

4.  Find  two  line-segments,  one  of  given  length,  which  shall  be  in  the 
same  ratio  as  the  areas  of  two  equilateral  triangles. 


328-330]  AREAS   OF  PLANE  POLYGONS  223 

Proposition  XIII 

329.   To  construct  a  triangle  equal  in  area  to  a  given 
triangle  and  having  one  side  of  given  length. 

M  N 


B  b  0  B'        b'  G' 

It  is  required  to  construct  on  the  line-segment  JB'C  a  triangle 
equal  in  area  to  a  given  triangle  ABC. 

Construction.  Let  h  and  h  be  the  base  and  altitude,  respec- 
tively, of  the  given  triangle,  and  6'  and  ^'  the  base  and  altitude 
of  the  required  triangle,  of  which  h'  is  unknown. 

Then  \lib  =  \ h'b',  or  hb  =  h'b'.  (Art.  306.) 

Therefore  b':b  =  k:h'; 

and  h'  can  be  determined  as  the  fourth  proportional  to  6', 
6,  and  h.  (Art.  272.) 

When  h'  has  been  so  determined,  draw  a  line  MN  parallel  to 
B'C,  whose  distance  from  B'C  equals  h'. 

Then  any  triangle  having  B'C  for  base  and  its  vertex  in 
MN  will  satisfy  the  given  conditions. 

Proposition  XIV 

330.  To  construct  a  triangle  equal  in  area  to  the  sum 
of  two  given  triangles  and  having  one  side  of  given 
length. 

1.  Construct  two  triangles,  P  and  Q,  equal  in  area  to  the 
given  triangles,  each  having  one  side  (the  base)  of  the  given 
length.  (Prop.  XIII.) 

2.  Construct  a  triangle,  R,  having  its  base  of  the  given 
length  and  its  altitude  equal  to  the  sum  of  the  altitudes  of 
P  and  Q. 

The  triangle  R  fulfils  the  given  conditions.     Prove. 


224 


ELEMENTARY  GEOMETRY 


[Chap.  IV 


Proposition  XV 

331.   To  reduce  the  number  of  sides  of  a  polygon  with- 
out altering  the  measure  of  its  area. 


Let  ABODE  be  the  given  polygon  in  which  A,  B,  and  C  are 
consecutive  vertices. 

Join  AC  and  through  B  dravr  BC  parallel  to  AC,  meeting 
the  line  DC  at  C.     Join  AC. 

'    AAC'C  =  AABC.  Why? 

Therefore  the  polygon  AC'DE  is  equal  in  area  to  the  origi- 
nal polygon  ABCDE,  and  it  has  one  less  side. 

By  repeated  applications  of  this  process,  the  number  of 
sides  of  any  polygon  can  be  reduced  to  three  without  altering 
the  area. 

EXERCISES 

1.  Apply  the  method  of  Proposition  XV  to  reduce  the  number  of 
sides  of  a  polygon  having  a  reentrant  angle,  as  in  the  dia- 
gram, the  area  to  remain  unaltered. 

2.  Construct  a  square  equal  in  area  to  a  given  polygon. 

First  apply  Proposition  XV  to  reduce  the  polygon 
to  a  triangle. 

3.  Construct  a  polygon  similar  to  a  given  polygon,  P,  and  equal  in 
area  to  another  given  polygon,  Q. 

Find  squares  equal  in  area  to  P  and  Q.  Let  m  and  n  be  sides 
of  these.  Let  a  be  any  side  of  P,  and  find  a'  a  fourth  pro- 
portional to  m,  w,  and  a.  On  a'  construct  a  polygon,  P, 
similar  to  P,  a  and  a'  being  homologous  sides.  R  is  the 
required  polygon.     Prove. 


331-333]  AEEAS   OF  PLANE  POLYGONS  .  225 


Proposition  XVI 

332.   To  construct  a  rectangle  having  given  its  area 
and  the  sum  of  two  adjacent  sides,  or  its  perimeter. 


A      F  B 

Let  P  be  equal  to  the  given  area  and  AB  the  sum  of  two 
adjacent  sides. 

It  is  required  to  construct  a  rectangle  having  its  area  equal 
to  P  and  its  perimeter  equal  to  twice  AB. 

Construction.  Find  m  a  side  of  the  square  whose  area  is 
equal  to  P.  (Ex.  2,  p.  224.) 

On  AB  describe  a  semicircle. 

At  A  erect  a  perpendicular  AC  equal  to  m,  and  through 
C  draw  a  straight  line  parallel  to  AB,  cutting  the  semicircle 
at  D  and  E. 

Draw  DF  perpendicular  to  AB,  and  join  AD  and  BD. 

DF  =  m,  and  DF'  =  AF'FB=  m\  (Art.  255.) 

Therefore  the  rectangle  whose  adjacent  sides  are  AF  and 
FB  is  the  one  required. 

333.  In  the  above  proposition  suppose  that  AB  remains  unaltered 
while  P,  the  given  area,  is  chosen  larger.  Then  m,  the  side  of  the  equiva- 
lent square,  becomes  longer  and  may  be  so  long  that  the  line  drawn 
through  C  parallel  to  AB  will  not  cut  the  semicircle.  In  that  case,  no 
rectangle  can  be  constructed  satisfying  the  conditions. 

From  a  consideration  of  the  conditions  of  this  problem  answers  to  the 
following  questions  will  readily  present  themselves  : 

1.  What  is  the  greatest  length  of  m  for  which  the  required  rectangle 
can  be  constructed  ? 

2.  What  rectangle  of  given  perimeter  will  have  the  greatest  area  ? 

3.  What  parallelogram  with  a  given  base  and  given  perimeter  will 
have  the  greatest  area  ? 

Such  questions  naturally  lead  to  a  further  consideration  of  what  we 
have  called  maximum  and  minimum  values.     (See  Art.  175.) 
Q 


226  ELEMENTARY  GEOMETRY  [Chap.  IV 

Section  IV 

MAXIMA  AND   MINIMA 

334.  When  a  geometrical  figure  varies  continuously  it  some- 
times happens  that  one  or  more  of  the  variable  magnitudes 
connected  with  it  gradually  increases  for  a  time  and  afterward 
begins  to  decrease.  When  such  a  magnitude  has  reached  its 
greatest  value,  that  is,  just  when  it  ceases  to  increase  and 
begins  to  decrease,  it  is  said  to  have  a  maximum  value.  On 
the  other  hand,  if  the  magnitude  is  first  decreasing  and  then 
begins  to  increase,  at  the  time  the  change  takes  place  the 
magnitude  is  said  to  have  a  minimum  value. 

Suppose  we  recall  the  example  given  in  Article  175. 

Take  within  a  circle  any  point  P,  not  the  centre,  and  join  it  to  a  point 
A  on  the  circle.  Let  A  move  continuously  along  the 
circle  in  the  way  indicated  by  the  arrowhead.  The 
straight  line  PA  will  steadily  increase  in  length  till 
it  reaches  that  position  in  which  it  passes  through 
the  centre  ;  after  that  it  will  steadily  decrease  for  a 
time. 

So  we  say,  the  line  PA  is  a  maximum  when  it 
passes  through  the  centre. 

As  the  rotation  continues,  the  line  PA  will  reach 
a  minimum  in  that  position  where,  if  produced  backward,  it  would  pass 
through  the  centre. 

Since  a  maximum  or  minimum  value  of  a  variable  magnitude 
occurs  at  a  critical  position  which  may  be  approached  from 
either  side,  we  naturally  look  for  a  maximum  or  a  minimum 
when  the  variable  magnitude  is  symmetrically  situated  with 
respect  to  the  rest  of  the  figure,  and  this  usually  proves  to  be 
the  case. 

Thus,  the  shortest  straight  line  from  a  given  point  to  a 
straight  line  not  passing  through  it  is  the  perpendicular  to 
the  line. 


334] 


AREAS   OF  PLANE  POLYGONS 


227 


Examples  of  Maxima  and  Minima 

1.  If  two  sides  of  a  triangle  are  of  given  lengths  while  the 
third  side  varies,  the  triangle  ivill  have  the  maximum  area 
when  the  first  two  sides  are  at  right  angles. 

Let  AB  and  AC  be  given  sides  of  a  triangle  ABC,  and  let 
the  angle  between  them  vary  con- 
tinuously, so  that,  while  AB  re- 
mains fixed,  the  side  AC  rotates 
about  A,  taking  up  successively 
the  positions  AC^,  AC 2,  AC^, .  .  . 

The  area  of  A  ABC  equals  one- 
half  the  product  of  AB  and  the 

perpendicular  from  C  upon  AB.  The  perpendicular  CD  steadily 
increases  with  the  rotation  until  AC  makes  right  angles  with 
AB,  after  which  it  begins  to  decrease. 

The  area  of  A  ABC  therefore  gradually  increases  with  the  ro- 
tation till  ^C  is  perpendicular  to  AB,  after  which  it  decreases. 

Hence  the  area  is  a  maximum  when  AC  and  AB  are  at 
right  angles. 

2.  In  a  straight  line,  to  find  a  point  such  that  the  sum  of  its 
distances  from  tivo  given  points  on  the  same  side  of  the  line  shall 
be  a  minimum. 

G  D 


<2\. 


D 


The  problem  is  to  find  a  point  Q  in  AB  such  that  CQ  -f-  QD 
has  a  minimum  value. 

Let  D'  be  the  inverse  point  (Art.  144)  of  D  with  respect  to 
the  line  AB.  Take  any  point  Q  in  AB.  Then  QD'  =  QD. 
Why? 


228 


ELEMENTARY  GEOMETRY 


[Chap.  IV 


O 


Therefore  CQ  +  QD'  =  CQ  +  QD. 

Hence  CQ  -\-  QD  is  least  when  CQ  +  QD'  is  least. 

What  then  is  the  required  position  of.  Q? 

3.  Given  two  intersecting  straight  lines  AB,  AC,  and  a  point  P 
between  them;  of  all  line-segments  ichich  pass  through  P  and  are 
terminated  by  AB  and  AC,  that  which  is  bisected  at  P  makes  with 
AB  and  AC  the  triangle  of  minimum  area. 

Let  the  line-segment  DE  be  bisected  at  P.  It  is  required  to 
show  that  A  DAE  is  less  than  A  FAG,  where  FG  is  any  other 
line-segment  through  P. 

Through   D  draw   DM  parallel 
to^C. 

Then  A  PDM  is  identically  equal 
to  A  PEG.     Why  ? 

But  A  PDM  is  less  than  A  PDF. 

Therefore  A  PEG  is   less   than 
A  PDF. 

To  each  add  the  figure  ADPG. 

Therefore  A  DAE  is  less  than 
A  FAG. 

EXERCISES 

1.  What  is  the  maximum  chord  of  a  circle  ?  What  is  the  minimum  line- 
segment  from  a  given  point  outside  to  a  given  circle  ? 

2.  Of  all  triangles  having  the  same  base  and  equal  areas,  that  which  is 
isosceles  has  the  minimum  perimeter. 

8.  Of  all  triangles  having  the  same  area,  that  which  is  equilateral  has 
the  minimum  perimeter. 

4.  Of  all  triangles  having  the  same  base  and  equal  perimeters,  that 
which  is  isosceles  has  the  maximum  area. 

Definition.    Figures  having  equal  perimeters  are  called  isoperimetric. 

5.  Given  the  base  and  the  vertical  angle  of  a  triangle,  construct  the 
triangle  so  that  its  area  may  be  a  maximum. 

6.  Divide  a  given  line-segment  into  two  parts  so  that  the  sum  of  the 
squares  on  the  parts  may  be  a  minimum. 

7.  Divide  a  given  line-segment  into  two  parts  so  that  the  rectangle 
contained  by  the  parts  is  a  maximum. 


E 

/ 

a/ 

/\ 

P 

X 

D 


334]  AREAS  OF  PLANE  POLYGONS  229 


MISCELLANEOUS    EXERCISES 

1.  Construct  a  parallelogram  equal  in  area  to  a  given  parallelogram, 
and  having  one  of  its  angles  equal  to  a  given  angle. 

2.  ABC  is  a  given  triangle  ;  construct  a  triangle  of  equal  area  having 
AB  for  base  and  its  vertex  in  a  given  straight  line. 

3.  ABCD  is  a  parallelogram  ;  from  any  point  P  in  the  diagonal  BD^ 
the  straight  lines  P^l,  PC  are  drawn.  Show  that  the  triangles  PAB  and 
PCB  are  equal  in  area. 

4.  The  sides  AB,  AC  ot  a.  triangle  are  bisected  at  C'B\  respectively  ; 
CC,  BB'  intersect  at  F.  Prove  that  the  triangle  BFC  is  equal  to  the 
quadrilateral  AC'FB'. 

5.  The  locus  of  a  point  such  that  the  sum  of  the  squares  on  its  dis- 
tances from  two  fixed  points  is  equal  to  the  square  on  the  distance  between 
the  points  is  a  circle  passing  through  the  two  points. 

6.  The  locus  of  a  point  such  that  the  difference  of  the  squares  on  its 
distances  from  two  fixed  points  is  equal  to  the  square  on  the  distance 
between  the  points  is  a  straight  line. 

7.  If  ABCD  is  a  square  and  the  vertex  A  is  joined  to  the  mid-point  of 
the  side  BC,  B  to  the  mid-point  of  CD,  C  to  the  mid-point  of  DA,  and  D 
to  the  mid-point  of  AB,  the  lines  so  drawn  are  the  sides  of  a  square  whose 
area  is  one-fifth  of  the  area  of  the  original  square. 

8.  Construct  two  line-segments  in  the  ratio  1 :  V2. 

9.  Divide  a  given  triangle  into  two  equal  parts  by  a  straight  line 
parallel  to  one  of  its  sides. 

10.  Construct  two  line-segments  in  the  ratio  1 :  VS. 

11.  Draw  a  straight  line-parallel  to  one  side  of  a  given  triangle  so  as  to 
form  a  triangle  equal  to  one-third  of  the  original  triangle. 

12.  If  three  line-segments  are  in  continued  proportion,  the  first  is  to 
the  third  as  the  area  of  a  triangle  described  on  the  first  is  to  the  area  of 
the  similar  triangle  described  on  the  second. 

13.  If  any  point  within  a  parallelogram  is  joined  to  the  four  vertices, 
the  sum  of  either  pair  of  triangles  having  parallel  bases  is  half  the  area  of 
the  parallelogram. 

14.  The  sides  of  a  triangle  are  21,  24,  35  feet,  respectively.  Find  the 
area  of  the  triangle  and  the  projection  of  the  lesser  side  upon  each  of  the 
others. 


230  ELEMENTARY  GEOMETRY  [Chap.  IV 

SUMMARY  OF   CHAPTER  IV 

1.  Definitions. 

(1)  Adjacent  Polygons  —  polygons  having  one  or  more  sides  or  parts 

of  sides  in  common.     §  288. 

(2)  Sum  of  Two  Polygons  —  a  polygon  formed  by  making  the  two 

adjacent  and  disregarding  the  common  boundary.     §  288. 

(3)  Area  of  a  Plane  Closed  Figure  —  surface  enclosed  by  the  figure, 

or  the  measure  of  that  surface.     §§  289,  291. 

(4)  Altitude  of  a  Parallelogram  —  the  distance  between  the  base  and 

the  opposite  side.     §  298. 

(5)  Altitude  of  a  Triangle. —  the  distance  between  the  base  and  the 

opposite  vertex.     §  298. 

(6)  Projection  of  a  Point  on  a  Line  —  the  foot  of  the  perpendicular 

drawn  from  the  point  to  the  line.     §  319. 

(7)  Projection  of  a  Given  Line-segment  on  Any  Straight  Line  —  the 

intercept  between  the  projections  of  the  extremities  of  the 
segment  on  the  line.     §  319. 

(8)  Isoperimetric  Figures  —  those  having  equal  perimeters.  Ex.  4, 

p.  228. 

2.  Axioms. 

(1)  If  two  plane  polygons  or  other  closed  figures  are  identically 

equal,  their  areas  are  equal  (Axiom  11).     §  289. 

(2)  The  sum  of  the  areas  of  two  plane  polygons  is  equal  to  the  area 

of  their  sum  (Axiom  12).     §  289. 

3.  Problems. 

(1)  To  construct  a  rectangle  equal  in  area  to  a  given  parallelogram, 

and  having  one  side  equal  to  a  given  line-segment.     §  325. 

(2)  To  construct  a  square  equal  in  area  to  the  sum  of  two  given 

squares.     §  327. 

(3)  To  find  two  line-segments  having  the  same  ratio  as  the  areas  of 

two  given  squares.     §  328. 

(4)  To  construct  a  triangle  equal  in  area  to  a  given  triangle  and 

having  one  side  of  given  length.     §  329. 

(5)  To  construct  a  triangle  equal  in  area  to  the  sum  of  two  given 

triangles  and  having  one  side  of  given  length.     §  330. 

(6)  To  reduce  the  number  of  sides  of  a  polygon  without  altering  the 

measure  of  its  area.     §  331. 

(7)  To  construct  a  rectangle  having  given  its  area  and  the  sum  of  two 

adjacent  sides,  or  its  perimeter.    §  332. 


Summary]  AREAS   OF  PLANE  POLYGONS  231 

(8)  In  a  straight  line  to  find  a  point  such  that  the  sum  of  its  distances 
from  two  given  points  on  the  same  side  of  the  line  shall  be  a 
minimum.     §  334.  2. 

4.  Theorems  on  the  Areas  of  Parallelograms. 

(1)  Parallelograms  upon  the  same  base  and  between  the  same  par- 

allels are  equal  in  area.     §  292. 

(2)  Parallelograms  upon  equal  bases  and  between  the  same  parallels 

are  equal  in  area.     §  294. 

(3)  Any  parallelogram  is  equal  in  area  to  a  rectangle  having  an  equal 

base  and  an  equal  altitude.     §  299. 

(4)  The  areas  of  two  rectangles  having  equal  altitudes  are  in  the 

same  ratio  as  their  bases.    §  300. 

(5)  The  areas  of  any  two  rectangles  are  in  the  same  ratio  as  the 

products  of  their  bases  and  altitudes.     §  303. 

(6)  The  area  of  any  rectangle  is  equal  to  the  product  of  its  base  and 

its  altitude.     §  304. 

(7)  The  area  of  a  parallelogram  is  equal  to  the  product  of  its  base 

and  its  altitude.     §  305. 

(8)  The  area  of  a  square  is  equal  to  the  square  of  any  one  of  its 

sides.     §  307. 

(9)  The  areas  of  two  parallelograms  having  an  angle  of  the  one  equal 

to  an  angle  of  the  other  are  in  the  same  ratio  as  the  products 
of  the  sides  containing  the  equal  angles.     §  309. 

5.  Theorems  on  the  Areas  of  Triangles. 

(1)  If  a  triangle  and  a  parallelogram  are  upon  the  same  base,  or  upon 

equal  bases,  and  between  the  same  parallels,  the  ar^a  of  the 
triangle  equals  half  the  area  of  the  parallelogram.     §  295. 

(2)  Triangles  upon  the  same  base,  or  upon  equal  bases,  and  between 

the  same  parallels,  are  equal  in  area.     §  296. 

(3)  Triangles  upon  equal  bases  in  the  same  straight  line,  having  their 

opposite  vertices  in  common,  are  equal  in  area.     §  297. 

(4)  The  areas  of  two  triangles  having  equal  altitudes  are  in  the  same 

ratio  as  their  bases ;  or  having  equal  bases  are  in  the  same 
ratio  as  their  altitudes.     §  302. 

(5)  The  area  of  a  triangle  is  equal  to  half  the  product  of  its  base  and 

altitude.     §  306. 

(6)  The  areas  of  two  triangles  having  an  angle  of  the  one  equal  to  an 

angle  of  the  other  are  in  the  same  ratio  as  the  products  of  the 
sides  containing  the  equal  angles.     §  308. 

(7)  Area  of  a  triangle  in  terms  of  the  sides.    See  §  324. 


232  ELEMENTARY  GEOMETRY  [Chap.  IV 

6.  Theorems  on  the  Areas  of  Similar  Polygons. 

(1)  The  areas  of  similar  triangles  are  in  the  same  ratio  as  the  squares 

of  any  two  homologous  sides.     §  314. 

(2)  The  areas  of  similar  polygons  are  in  the  same  ratio  as  the  squares 

of  any  two  homologous  sides.     §  315. 

(3)  The  areas  of  similar  polygons  are  in  the  same  ratio  as  the  squares 

of  any  two  homologous  diagonals.     §  316. 

(4)  The  area  of  the  square  described  on  the  hypotenuse  of  a  right 

triangle  is  equal  to  the  sum  of  the  areas  of  the  squares  described 
on  the  other  two  sides.     §  317. 

(5)  The  area  of  any  polygon  described  on  the  hypotenuse  of  a  right 

triangle  is  equal  to  the  sum  of  the  areas  of  the  similar  polygons 
similarly  described  on  the  other  two  sides.     §  318. 

7.  Miscellaneous  Theorems. 

(1)  If  a  given  line-segment  is  divided  internally  into  any  two  parts, 

the  square  on  the  whole  segment  is  equal  in  area  to  the  sum  of 
the  squares  on  the  two  parts  together  with  twice  the  rectangle 
contained  by  the  two  parts.     §  311. 

(2)  If  a  given  line-segment  is  divided  externally  into  any  two  parts, 

the  square  on  the  given  segment  is  equal  in  area  to  the  sum  of 
the  squares  on  the  two  parts  less  twice  the  rectangle  contained 
by  the  two  parts.     §  312. 

(3)  In  any  triangle  the  square  on  the  side  opposite  an  acute  angle  is 

less  than  the  sum  of  the  squares  on  the  other  two  sides  by 
twice  the  product  of  either  of  these  sides  and  the  projection  of 
the  other  upon  it.     §  321. 

(4)  In  an  obtuse-angled  triangle  the  square  on  the  side  opposite  the 

obtuse  angle  is  greater  than  the  sum  of  the  squares  on  the 
other  two  sides  by  twice  the  product  of  either  of  these  sides 
and  the  projection  of  the  other  upon  it.     §  322. 

(5)  If  two  sides  of  a  triangle  are  of  given  lengths  while  the  third  side 

varies,  the  triangle  will  have  the  maximum  area  when  the  first 
two  sides  are  at  right  angles.     §  334.  1. 

(6)  Given  two  intersecting  straight  lines  AB^  AC,  and  a  point  P 

between  them  ;  of  all  line-segments  which  pass  through  P  and 
are  terminated  by  AB  and  AC,  that  which  is  bisected  at  P 
makes  with  AB  and  ^(7  the  triangle  of  minimum  area.  §  334. 3. 


CHAPTER   V 
MEASUREMENT  OF  THE   CIRCLE 

Section  I 

REGULAR   POLYGONS 
Definitions 

335.  A  regular  polygon  is  a  polygon  which  is  both  equi- 
lateral and  equiangular. 

For  example,  an  equilateral  triangle  is  a  regular  polygon  of 
three  sides ;  a  square  is  a  regular  polygon  of  four  sides. 

A  polygon  of  more  than  three  sides  may  be  equilateral  with- 
out being  equiangular,  or  it  may  be  equiangular  without  being 
equilateral ;  but  in  order  to  be  classed  as  regular  it  must  be 
both  equilateral  and  equiangular. 

That  there  can  be  regular  polygons  of  any  given  number  of 
sides  will  be  seen  from  the  first  proposition  of  this  chapter. 

Make  a  diagram  of  a  quadrilateral  that  is  (1)  equilateral,  but  not  equi- 
angular;  (2)  equiangular,  but  not  equilateral;  (3)  both  equilateral  and 
equiangular  ;  (4)  neither  equilateral  nor  equiangular. 

336.  A  regular  polygon  of  more  than  four  sides  may  or  may 
not  be  convex,  but  unless  the  contrary  is  stated,  it  will  be 
understood  that  any  figure  under  discussion  is  convex. 

A  polygon  of  five  sides  is  called  a  pentagon  ;  one  of  six  sides, 
a  hexagon ;  one  of  seven  sides,  a  heptagon ;  one  of  eight  sides, 
an  octagon ;  one  of  ten  sides,  a  decagon ;  one  of  twelve  sides,  a 
dodecagon. 

233 


234  ELEMENTARY  GEOMETRY  [Chap.  V 

337.  Postulate  8.  A  circle  may  he  divided  into  any  given 
number  of  equal  arcs. 

The  problem  "  to  divide  a  circle,  or  any  arc  of  a  circle,  into  a 
given  number  of  equal  parts"  is  not  always  solvable  by  the 
methods  of  elementary  geometry.  The  method  of  solution,  if 
there  is  one,  must  of  course  depend  on  the  number  of  such 
parts  required.  All  that  the  above  postulate  affirms  is  that 
the  circle  may  be  thought  of  as  made  up  of  any  specified 
number  of  equal  parts,  and  the  points  of  division  may  be 
assumed.  It  says  nothing  at  all  about  a  method  of  finding 
these  points  of  division. 

338.  In  the  preceding  chapters  we  have  shown  how  to  divide 
line-segments,  angles,  and  arcs  of  circles,  into  certain  numbers 
of  equal  parts.  The  following  exercises  will  recall  the 
methods  employed. 

EXERCISES 

1.  Divide'  a  given  line-segment  into  two  equal  parts ;  into  four  equal 
parts ;  into  eight  equal  parts  (Art.  56). 

2.  Divide  a  given  line-segment  into  three  equal  parts ;  into  six  equal 
parts  ;  into  nine  equal  parts  (Art.  142) . 

3.  Divide  a  given  line-segment  into  any  required  number  of  equal 
parts  (Art.  269). 

4.  Divide  a  given  angle  into  two,  four,  eight,  etc.  equal  parts  (Art.  54). 

5.  Divide  a  right  angle  into  three  equal  parts  (Art.  142). 

6.  Divide  a  given  arc  of  a  circle  into  two,  four,  eight,  etc.,  equal  parts 
(Art.  165). 

The  problem  to  divide  any  given  angle  or  arc  of  a  circle  into 
three  equal  parts  has  been  found  to  be  impossible  of  solution 
by  the  methods  of  elementary  geometry,  in  which  we  make 
use  of  the  ruler  and  compasses  only. 

7.  Divide  a  circle  into  two,  four,  eight,  etc.,  equal  arcs. 

S'  Divide  a  circle  into  three,  six,  or  twelve  equal  arcs.  (Make  use  of 
Ex.  5.) 


337-339] 


MEASUREMENT  OF  THE  CIRCLE 


235 


Proposition  I 

339.  If  a  circle  is  divided  into  any  number  of  equal 
arcs: 

I.  The  chords  joining  the  points  of  division,  taken  in 
order,  form  a  regular  inscribed  polygon. 

II.  The  tangents  to  the  circle  at  the  points  of  division, 
tahen  in  order,  form  a  regular  circumscribed  polygon. 

Remakk.     The  number  of  sides  of  the  polygon  is  equal  in  each 
case  to  the  number  of  points  of  division  in  the  circle. 


Let  a  given  circle  be  divided  into  any  number  of  equal  arcs 
AB,  BC,  CD,  etc.,  and  let  the  points  of  division  be  joined  in 
order,  thus  forming  the  polygon  ABODE ;  also  at  the  points  of 
division,  let  the  tangents  to  the  circle  be  drawn,  thus  forming 
the  polygon  FGHJK. 

It  is  required  to  prove  that  the  polygon  ABCDE  is  a  regular 
polygon ;  and  also,  that  the  polygon  FGHJK  is  a  regular  poly- 
gon. 

Proof.  First,  since  by  hypothesis  the  arcs  AB,  BC,  CD,  etc., 
are  equal,  the  chords  AB,  BC,  CD,  etc.,  are  equal.    (Art.  161.) 

Therefore  the  polygon  ABCDE  is  equilateral. 

Also,  the  arc  EAB  =  the  arc  ABC     Why  ? 


Therefore,  Z  EAB  =  Z  ABC. 


(Art.  181.) 


236  ELEMENTARY  GEOMETRY  [Chap.  V 

Similarly  Z  BCD  =  Z  ABC,  and  so  on. 
Therefore  the  polygon  ABCDE  is  equiangular. 
That  is,  the  polygon  ABCDE  being  both  equilateral  and 
equiangular  is  regular. 


Next,  let  the  points  A,  B,  C,  etc.,  be  joined  to  the  centre  0. 
Then  Zs  AOB,  BOC,  COD,  etc.,  are  all  equal.     Why  ? 
Therefore,  Zs  AFB.  BOC,  CUD,  being  supplements  of  equal 
angles  at  the  centre  [why  ?]  are  also  equal. 
Hence  the  polygon  FOHJK  is  equiangular. 
Also,  the  line-segment  AF  =  the  line-segment  FB.   (Art.  194.) 
Also,  the  line-segment  FB  =  the  line-segment  BG.     Prove. 
Similarly  BG  =  GC,  GC  =  CH,  and  so  on. 
Therefore  the  side  FG  =  the  side  GH,  =  the  side  HJ,  and  so  on. 
Hence  the  polygon  FGHJK  is  equilateral. 
That  is,  the  polygon  FGHJK  is  also  regular. 

340.  Corollary  T.  If  the  vertices  of  a  regular  inscribed 
polygon  are  joined  to  the  mid-points  of  the  arcs  subtended  by  the 
sides  of  the  polygon,  the  joining  lines  will  form  another  regidar 
polygon  of  twice  the  number  of  sides. 

341.  Corollary  II.  If  tangents  are  drawn  at  the  mid-points 
of  the  arcs  between  the  points  of  contact  of  the  sides  of  a  regular 
circumscribed  polygon,  these  together  with  the  sides  of  the  original 
polygon  form  a  regular  polygon  of  twice  the  number  of  sides. 


340-342]  MEASUBEMENT  OF  THE  CIECLE  287 


Proposition  II 

342.  Any  equilateral  polygon  inscribed  in  a  circle  is 
also  equiangular,  and  hence  regular. 

For  the  equal  sides  of  the  polygon  subtend  equal  arcs  of  the 
circle  (Art.  163),  these  arcs  together  making  up  the  whole  circle 
(or  some  multiple  of  the  circle,  if  the  polygon  is  not  convex). 
The  angles  between  consecutive  sides  of  the  polygon  also  sub- 
tend equal  arcs,  and  are  therefore  equal.  Hence  the  polygon 
is  regular. 

Is  the  converse  of  this  theorem  true  ? 


EXERCISES 

1.  If  a  circle  is  divided  into  five  equal  arcs  at  the  points  A^  B,  C,  D,  E, 
and  the  points  are  joined,  each  to  the  next  but 
one,  as  in  the  diagram,  prove  that  the  result- 
ing figure  is  a  regular  polygon,  according  to 
the  definition,  though  not  a  convex  regular 
polygon. 

Suppose  the  circle  is  divided  into  seven 
equal  arcs,  construct  an  inscribed  polygon  by 
joining  each  vertex  to  the  next  but  one,  next 
but  two,  next  but  three,  etc.  Show  that  the 
polygons  so  constructed  are  regular. 

2.  All  regular  convex  polygons  of  the  same  number  of  sides  inscribed 
in  the  same  circle  are  equal ;  and  all  regular  convex  polygons  of  the 
same  number  of  sides  circumscribed  about  the  same  circle  are  equal. 

3.  The  perimeter  of  a  regular  polygon  inscribed  in  a  circle  is  less  than 
the  perimeter  of  a  circumscribed  polygon  of  the  same  number  of  sides. 

4.  The  perimeter  of  a  regular  polygon  inscribed  in  a  circle  is  less  than 
an  inscribed  polygon  of  twice  the  number  of  sides. 

5.  Any  parallelogram  inscribed  in  a  circle  is  rectangular. 

6.  If  a  rectangle  is  circumscribed  about  a  circle  it  must  be  a  square. 

7.  What  is  the  measure  of  the  angle  at  the  centre  of  a  circle  subtended 
by  the  side  of  a  circumscribed  square  ? 


238 


ELEMENTARY  GEOMETRY 


[Chap.  V 


Proposition  III 

343.   t^  circle  can  be  circumscribed  about  any  regular 
polygon,  and  another  circle  can  be  inscribed  in  it. 


B.^'-'  H 


Let  ABODE  ...  be  any  regular  polygon. 

It  is  required  to  prove  that  a  circle  can  be  circumscribed  about 
it,  and  that  another  circle  can  be  inscribed  in  it. 

Proof.  First,  at  the  mid-points  H  and  K  of  two  adjacent 
sides,  AB  and  BC,  of  the  given  polygon,  draw  the  perpendicu- 
lars to  these  sides,  and  let  them  meet  at  the  point  0. 

Then  0  is  the  centre  of  the  circle  which  can  be  made  to  pass 
through  the  three  vertices  A,  B,  and  C.  (Art.  151.) 

Since  OA  =  OB  =  00,  and  AB  =  BO,  As  OAB  and  OBO  are 
equal  isosceles  triangles,  and  therefore 

Z  OAB  =  Z  OBA  =  Z  OBO=Z  OOB, 

That  is,  Z  OBO  is  one  half  of  Z  ABO. 

Therefore  Z  OOB  is  one  half  of  the  equal  angle  BODy 

or  Z  OOB  =  Z  OOD. 

Join  OD. 
Then  As  OBO  and  ODO  are  identically  equal.         (Art.  41.) 


Therefore 
and 


OD  =  OB, 
Z  ODO=Z  0B0=\  Z  ODE. 


343-348]  MEASUREMENT  OF  THE  CIRCLE  239 

Similarly,  OE  may  be  shown  to  be  equal  to  OC,  and  so  on 
for  other  vertices. 

Therefore  the  circle  which  passes  through  the  vertices  A,  B, 
and  C  will  also  pass  through  the  remaining  vertices  Z>,  E,  etc. 

Hence,  a  circle  can  be  circumscribed  about  the  given  polygon. 

Next,  since  As  AOB,  BOC,  COD,  etc.,  are  equal  isosceles  tri- 
angles, their  altitudes  OH,  OK,  OL,  OM,  and  O^are  all  equal. 
Prove. 

Therefore  the  circle  described  with  centre  0,  and  radius  OH, 
will  pass  through  the  points  K,  L,  M,  N,  and  the  sides  of  the 
polygon  will  be  tangent  to  the  circle,  since  each  side  is  at  right 
angles  to  a  radius  at  its  extremity.  (Art.  187.) 

Therefore  a  circle  can  be  inscribed  in  the  given  polygon. 

Definitions 

344.  The  common  centre  of  the  inscribed  and  circumscribed 
circles  of  a  regular  polygon  is  called  the  centre  of  the  polygon. 

345.  The  radius  of  the  circumscribed  circle,  or  the  line- 
segment  joining  the  centre  and  a  vertex  of  a  regular  polygon,  is 
called  the  radius  of  the  polygon. 

346.  The  radius  of  the  inscribed  circle,  or  the  perpendicular 
from  the  centre  on  a  side  of  a  regular  polygon,  is  called  the 
apothem  of  the  polygon. 

347.  Corollary  I.  Aiiy  radius  of  a  regular  polygon  bisects 
the  angle  at  the  vertex. 

348.  Corollary  II.  The  angle  formed  by  two  consecutive 
radii  of  a  regular  polygon  equals  four  right  angles  divided  by  the 
number  of  sides  of  the  polygon. 

This  angle  is  sometimes  spoken  of  as  ^  the  angle  at  the  centre 
of  the  polygon.' 


240 


ELEMENTARY  GEOMETRY 


[Chap.  V 


Proposition  IV 

349.   Any  two  regular  polygons  of  the  same  numher 
of  sides  are  similar. 


4^'- 


Let  ABCDE  and  A'B'OD'E'  be  two  regular  polygons  of  the 
same  number  of  sides. 

It  is  required  to  prove  that  they  are  similar. 

Proof.  First,  the  two  polygons  are  mutually  equiangular, 
since  in  each  polygon  the  interior  angles  are  all  equal,  and  the 
magnitude  of  an  interior  angle  of  either  polygon  depends  only 
on  the  number  of  sides,  which  is  the  same  in  both. 

Next,  the  homologous  sides  are  proportional. 

For  AB  =  BC=CD  =  etc., 

and  A'B^  =  B'C  =  CD'  =  etc. 

Therefore  AB  :  A'B'  =  BC :  B'C  =  CD  :  CD'  =  etc. 
Therefore  the  two  polygons  are  similar. 


350.  Corollary  I.  Homologous  sides  in  two  regular  poly- 
gons of  the  same  number  of  sides  are  in  the  same  ratio  as  the 
radii  of  the  circumscribed  circles,  or  as  the  radii  of  the  inscribed 
circles;  that  is,  as  the  radii  of  the  polygons,  or  as  the  apothems  of 
the  polygons. 

Suggestion.  Draw  the  radii  and  apothems  and  so  obtain  similar  tri- 
angles in  the  two  polygons. 


349-352]  MEASUREMENT  OF  THE  CIRCLE  241 

351.  Corollary  II.  The  perimeters  of  two  regular  polygons 
of  the  same  number  of  sides  are  in  the  same  ratio  as  their  radiiy 
or  as  their  apothems. 

Let  S\  and  ^'2  be  the  lengths  of  the  sides  in  two  regular  polygons  of  n 
sides,  of  which  A\  and  Ao  are  the  apothems.  Then  8]_:  S^^Ai  :  Ai  (Cor.  I). 
Therefore  n8\ :  n8i  =  A\ :  A2,  that  is,  the  perimeters  are  in  the  same  ratio 
as  the  apothems.     The  same  holds  true  for  the  radii. 

352.  Corollary  III.  The  areas  of  two  regular  pyolygons  of 
the  same  number  of  sides  are  in  the  same  ratio  as  the  squares  of 
their  radii,  or  as  the  squares  of  their  apothems.  (Art.  315.) 

EXERCISES 

1.  Tangents  to  a  circle  at  the  mid-points  of  the  arcs  subtended  by  the 
sides  of  a  regular  inscribed  polygon  form  a  regular  circumscribed  polygon. 

The  sides  of  the  circumscribed  polygon  are 
parallel  to  the  sides  of  the  inscribed  polygon 
(Art.  192),  and  the  vertices  of  the  circumscribed 
lie  upon  the  radii,  produced,   of  the  inscribed. 

Suggestion.  To  show  that  the  vertex  P  lies 
on  the  radius  OA^  show  by  similar  triangles  that 
the  tangent  at  H  cuts  OA  in  the  same  ratio  as 
does  the  tangent  at  K. 

Are  these  two  polygons  similar  ? 

o 

2.  The  interior  angle  of  a  regular  polygon  of  n  sides  equals  -  (w  —  2) 

right  angles.     (Art.  118.)  '* 

3.  The  interior  angle  of  any  regular  polygon  is  the  supplement  of  the 

angle  at  the  centre  of  the  polygon ;  that  is,  it  is  equal  to  180°—  — . 

n 
Show  that  this  statement  agrees  with  the  statement  of  Ex.  2. 

4.  Prove  that  the  side  of  a  regular  hexagon  is  equal  to  a  radius  of  the 
circumscribed  circle. 

5.  Find  in  degrees  the  angle  (1)  of  a  regular  pentagon,  (2)  of  a  regular 
hexagon,  (3)  of  a  regular  octagon. 

6.  Kegular  pentagons  are  inscribed  in  two  circles  of  five  and  eight  feet 
radius,  respectively.    Find  the  ratio  of  their  perimeters  and  of  their  areas. 


242  ELEMENTARY  GEOMETRY  [Chap.  V 


Proposition  V 

353.  The  area  of  a  regular  polygon  is  equal  to  half 
the  product  of  its  perimeter  and  its  apothem. 

Proof.  If  all  the  vertices  of  the  polygon  are  joined  to  the 
centre,  the  triangles  so  formed  are  all  equal,  and  the  area  of 
each  equals  half  the  product  of  a  side  and  the  apothem. 

(Art.  306.) 

Therefore  the  area  of  the  whole  polygon  equals  half  the 
product  of  its  perimeter  and  its  apothem. 


Proposition  VI 

354.  If  the  numher  of  sides  of  a  regular  inscribed 
polygon  be  doubled  the  perimeter  will  be  increased,  but 
if  the  nujnber  of  sides  of  a  regular  circumscribed  poly- 
gon be  doubled  the  perimeter  will  be  diminished. 


Let  ABODE  be  any  regular  polygon  inscribed  in  a  circle,  and 
FGHJK  a  regular  polygon  circumscribed  about  the  same  circle. 

For  convenience,  we  may  let  the  two  polygons  have  the 
same  number  of  sides  and  be  so  arranged  that  the  vertices  of 
the  inscribed  coincide  with  the  points  of  contact  of  the 
circumscribed. 


363-355]  MEASUREMENT  OF  THE  CIRCLE  243 

It  is  required  to  prove  that  if  the  number  of  sides  of  the 
inscribed  polygon  be  doubled,  say  by  joining  the  mid-points  of 
the  arcs  of  the  circle  to  the  vertices,  the  perimeter  will  be 
increased;  but  if  the  number  of  sides  of  the  circumscribed 
polygon  be  doubled,  say  by  drawing  tangents  at  the  mid-points 
of  the  arcs  of  the  circle,  the  perimeter  will  be  diminished. 

Proof.  Let  L  be  the  mid-point  of  the  arc  AE,  and  PQ  be 
tangent  to  the  circle  at  L. 

First,  in  A  ALE,  the  sum  of  AL  and  LE  is  greater  than 
AE.  (Art.  70.) 

Therefore  the  perimeter  of  the  regular  polygon  of  which  AL 
and  LE  are  two  adjacent  sides  is  greater  than  the  perimeter  of 
the  polygon  of  which  AE  is  a  side. 

Next,  in  A  FPQ,  the  side  FQ  is  less  than  the  sum  of  PF  and 
FQ. 

Therefore  the  sum  of  AP,  PQ,  and  QE  is  less  than  the  sum 
oiAFsiudFE. 

Hence  the  perimeter  of  the  regular  polygon  of  which  PQ  is 
one  side  is  less  than  the  perimeter  of  the  given  circumscribed 
polygon. 

Proposition  VII 

355.  The  area  of  a  regular  inscribed  polygon  is  in- 
creased, and  the  area  of  a  regular  circumscribed  polygon 
is  diminished,  when  the  number  of  sides  is  doubled. 

Proof.  The  area  of  a  regular  polygon  equals  half  the  product 
of  its  perimeter  and  its  apothem.  (Prop.  V.) 

When  the  number  of  sides  of  a  regular  inscribed  polygon  is 
doubled,  the  perimeter  is  increased  (Prop.  VI),  as  is  also  the 
apothem  [why  ?].     Hence  the  area  is  increased. 

When  the  number  of  sides  of  a  regular  circumscribed  polygon 
is  doubled,  the  perimeter  is  diminished  (Prop.  VI),  while  the 
apothem  remains  unaltered.     Hence  the  area  is  diminished. 


244  ELEMENTARY  GEOMETRY  [Chap.  V 

Section  II 

MEASUREMENT  OF  THE  CIRCLE 

356.  In  the  preceding  chapters,  whenever  we  have  spoken 
of  a  length,  we  have  had  in  mind  a  straight  line  distance ;  it 
has  been  in  every  case  the  measure  of  a  line-seginent. 

What  is  meant  by  the  length  of  a  curved  line  is  not  so  evi- 
dent, and  hence  it  is  necessary  to  give  this  expression  a  mean- 
ing by  definition, 

357.  If  any  regular  polygon  is  inscribed  in  a  given  circle, 
and  the  number  of  its  sides  is  repeatedly 

doubled  (or  is  indefinitely  increased  in 
any  regular  way),  the  polygon  can  be 
made  as  nearly  as  you  please  to  coincide 
with  the  circle. 

In  other  words,  the  circle  is  the  limit 
which  the  regular  inscribed  polygon  ap- 
proaches, as  the  number  of  its  sides  is 
indefinitely  increased. 

The  apothem  of  the  inscribed  polygon  approaches  the  radius 
of  the  circle  as  its  limit. 

358.  Definition.  The  length  of  a  circle  is  defined  to  be 
the  limit  of  the  perimeter  of  an  inscribed  regular  polygon,  as 
the  number  of  sides  of  the  polygon  is  indefinitely  increased. 

The  fact  that  the  perimeter  of  a  variable  inscribed  regular  polygon  has 
a  limit  admits  of  a  formal  proof,  the  essential  points  of  which  are : 
(1)  the  series  of  perimeters  is  constantly  increasing  (Prop.  VI);  (2)  the 
perimeter  never  exceeds  a  fixed  finite  quantity,  for  example,  the  perime- 
ter of  a  particular  circumscribed  polygon. 

The  length  of  a  circle  is  called  its  circumference. 

When  the  number  of  sides  of  an  inscribed  regular  polygon  is  indefi- 
nitely increased,  the  '  circle '  is  the  limit  of  the  '  polygon '  ;  the  '  circum- 
ference of  the  circle '  is  the  limit  of  the  '  perimeter  of  the  polygon.' 


356-361]  MEASUREMENT  OF  THE  CIRCLE  245 

The  length  of  an  arc  of  a  circle  is  defined  in  the  same  way 
to  be  the  limit  of  the  sum  of  chords  in  the  arc  when  the  num- 
ber of  such  chords  is  indefinitely  increased  in  some  regular 
way. 

359.  Definition.  The  area  of  a  circle  is  the  surface  en- 
closed by  the  circle.  It  is  equal  to  the  limit  of  the  area  of 
a  regular  inscribed   polygon   as  the 

number   of   its   sides   is  indefinitely 
increased. 

360.  The  circumference  of  a  circle 
could  just  as  well  be  defined  as  the 
limit  of  the  perimeter  of  a  regular 
circumscribed  polygon  when  the  num- 
ber of  its  sides  is  indefinitely  in- 
creased, since  the  limit  of  the  perim- 
eter of  the  circumscribed  polygon  is  the  same  as  the  limit  of 
the  perimeter  of  the  inscribed  polygon. 

This  statement  again  admits  of  a  formal  proof,  which  however  involves 
a  greater  knowledge  of  algebra  than  the  pupil  is  supposed  to  have  at  this 
stage.  The  assertions  that  the  circle  is  the  limit  of  the  regular  inscribed 
polygon  (Art.  357),  and  also  of  the  regular  circumscribed  polygon  (Art. 
360)  may  be  taken  as  postulates. 

The  radius  of  the  circumscribed  polygon  approaches  the 
radius  of  the  circle  as  its  limit,  and  the  area  of  the  polygon 
the  area  of  the  circle  as  its  limit. 

361.  Since  the  perimeter  of  the  inscribed  polygon  continually 
increases  and  the  perimeter  of  the  circumscribed  polygon  con- 
tinually decreases  as  the  number  of  their  sides  is  indefinitely 
increased  (Prop.  VI) : 

1.  The  circumference  of  a  circle  is  greater  than  the  perimeter 
of  any  regular  polygon  inscribed  in  it. 

2.  The  circumference  of  a  circle  is  less  than  the  perimeter  of 
any  regular  polygon  circumscribed  about  it. 


246 


ELEMENTARY  GEOMETRY 


[Chap.  V 


Pboposition  VIII 

362.   The  ratio  of  the  circumference  of  a  circle  to  its 
diameter  is  the  same  for  all  circles. 


Let  0  and  0'  be  the  centres  of  any  two  circles  whose  circum- 
ferences are  denoted  by  C  and  C,  their  diameters  by  d  and  d\ 
and  their  radii  by  r  and  r'. 

It  is  required  to  prove  that 

C:d  =  G':d'. 

Proof.  Inscribe  in  the  two  given  circles  regular  polygons 
ABODE  and  A'B'C'B'E'  of  the  same  number  of  sides.  The 
perimeters  of  these  polygons  are  in  the  same  ratio  as  their 
radii,  i.e.  as  the  radii  of  the  circles  in  which  they  are 
inscribed.  (Prop.  IV,  Cor.  II.) 

Let  P  and  P'  be  the  perimeters  of  the  two  polygons. 

Then  P:P'  =  r:r', 

P     P 

or  _  —  — . 


Suppose  now  the  number  of  sides  in  each  polygon  is  doubled, 
and  let  the  perimeters  of  the  polygons  so  formed  be  denoted 
by  Pi  and  P/. 

Then  ?1  =  El. 

r       r' 

If  this  process  is  repeated  indefinitely,  the  perimeter  P  will 
approach  the  circumference  C  as  its  limit,  and  the  perimeter 
P'  will  approach  the  circumference  O  as  its  limit.     (Art.  357.) 


362-363]  MEASUREMENT  OF  THE  CIRCLE  247 

(Art.  230.) 


Therefore 

l^          Ky 

7-r^- 

Also 

c     a 

2r  =  2r" 

c  a 
d    d'' 

or 


Since  the  two  given  circles  are  any  circles  whatsoever,  the 
ratio  of  the  circumference  to  the  diameter  in  any  one  circle  is 
equal  to  the  ratio  of  the  circumference  to  the  diameter  in  any 
other  circle. 

This  ratio  is  denoted  by  the  symbol  tt  (called  Pi),  so  that  for 
all  circles 

C 

—  =  TT, 

d       ' 
or  C  =  7rd  =  2  ttt. 

363.  It  should  be  carefully  noted  that  the  symbol  tt  represents  a  fixed 
and  definite  number,  whose  relation  to  other  numbers  however  we  have 
not  yet  determined.  The  old  problem  of  'squaring  the  circle,'  upon 
•which  so  much  time  and  labor  were  expended,  was  "to  find  the  side 
of  a  square  whose  area  is  equal  to  the  area  of  a  given  circle,"  and  this 
involved  finding  the  value  of  ir  in  terms  of  ordinary  numbers.  It  was 
shown  in  1882,  by  Lindemann,  a  German  mathematician,  that  ir  cannot 
be  so  expressed.  "We  can,  however,  approximate  its  value  in  ordinary 
numbers,  and  this  will  be  done  in  a  subsequent  proposition. 


EXERCISES 

1.  Show  that  the  area  of  an  inscribed  square  equals  2  r^. 

2.  Show  that  the  area  of  an  inscribed  regular  hexagon  equals  Sr^— -, 

,—  ^ 

and  that  the  distance  of  any  side  from  the  centre  is  r-^- 

3.  The  ratio  of  the  circumference  of  a  circle  to  the  perimeter  of  an 
inscribed  square  is  ir  :  2\/2,  and  to  the  perimeter  of  an  inscribed  hexagon 
is  IT  :  3. 


248 


ELEMENTARY  GEOMETRY 


[Chap.  V 


Proposition    IX 

364.   The  area  of  a  circle  is  equal  to  one-half  the 
product  of  its  circumference  and  radius. 


Let  A  be  the  area  of  the  given  circle,  r  its  radius,  and  C  its 
circumference. 

It  is  required  to  prove  that  A  =  \Cr. 

Proof.     Circumscribe  a  regular  polygon  about  the  given  circle, 
and  denote  its  perimeter  by  P,  and  its  area  by  H. 
Then  since  the  apothem  of  this  polygon  is  r, 


H 


1  Py. 

2  ^f- 


(Prop.  V.) 


If  now  the  number  of  sides  of  the  polygon  be  doubled,  and 
the  perimeter  of  this  new  polygon  be  denoted  by  Pi  and  its 
area  by  Hi,  then  again 

If  this  process  is  repeated  indefinitely,  the  ^perimeter  P 
approaches  the  circumference  C  as  its  limit,  iPr  approaches 
i  Cr  as  its  limit,  and  the  area  H  approaches  the  area  of  the 
circle  A  as  its  limit.  (Art.  360.) 

Since  the  variable  H  always  equals  the  variable  i  Pr,  the 
limit  A  equals  the  limit  ^  Cr.  (Art.  230.) 


Or 


A  =  iCr. 


364-368]  MEASUREMENT  OF  THE  CIRCLE  249 

365.   Corollary  I.     The  area  of  a  circle  is  equal  to  tt  times 
the  square  of  its  radius,  or  -  times  the  square  of  its  diameter. 

For  since  C  =  2  Trr  =  7rd, 

where  d  is  the  diameter  of  the  circle. 


366.  Corollary  II.  Tlie  areas  of  two  circles  are  in  the  same 
ratio  as  the  squares  of  their  radii,  or  as  the  squares  of  their 
diameters. 

If  A  is  the  area  of  a  circle  whose  radius  is  r  and  A'  is  the  area  of  a 
circle  whose  radius  is  r', 

and  A  =  il^  =  ^. 

A'      \ird'-^      d'-^ 


Definitions 

367.  Similar  arcs  of  circles   are   arcs    which   subtend  equal 
angles  at  the   centres  of 
the  circles  of  which  they 
are  parts. 

Thus  AB  and  A'B'  are 
similar  arcs  of  the  circles 
0  and  0',  if  the  angles  at 
the  centres  0  and  0'  are  equah 

368.  Similar  segments  of  circles   are   segments  which   have 
similar  arcs. 

Show  that  the  angles  inscribed  in  similar  arcs  or  similar  segments  are 
equal. 


250  ELEMENTARY  GEOMETRY  [Chap.  V 

369.  Similar  sectors  of  circles  are  sectors  which  have  similar 
arcs,  or  which  have  equal  angles  between  their  radii. 

370.  Theorem.  The  area  of  a  sector  of  a  circle  bears  the 
same  ratio  to  the  area  of  the  circle  as  the  length  of  its  arc  hears  to 
the  whole  circumference. 

371.  Corollary  I.  The  area  of  a  sector  of  a  circle  equals 
one-half  the  product  of  its  arc  and  radius. 

Let  the  arc  of  a  sector  equal  a  in  a  circle  of  radius  r,  and  denote  the 
area  of  the  sector  by  S. 

Then  S :  irr^  =  a  :  2  vr, 

that  is  S  =  I  ar. 

372.  Corollary  II.  The  arcs  of  similar  sectors  are  in  the 
same  ratio  as  the  radii,  and  the  areas  of  similar  sectors  are  in 
the  same  ratio  as  the  squares  of  the  radii. 

EXERCISES 

1.  The  area  of  one  circle  is  four  times  that  of  another.  Show  that  the 
radius  of  the  first  is  double  that  of  the  second. 

2.  The  diameter  of  one  circle  is  three  times  that  of  another  ;  compare 
their  circumferences  and  their  areas. 

3.  A  pond  is  fifty  yards  in  diameter ;  what  is  the  diameter  of  a  pond 
having  half  the  area  ? 

4.  The  arc  of  a  sector  of  a  circle  equals  the  radius  ;  show  that  the  area 
of  the  sector  is  to  the  area  of  the  circle  in  the  ratio  1 :  2  tt. 

5.  If  the  radius  of  a  circle  is  10  inches,  find  the  length  of  a  side  of  an 
inscribed  equilateral  triangle, 

6.  The  area  of  the  surface  between  two  concentric  circles  is  equal  to 
twice  the  area  of  the  smaller  circle.    Find  the  ratio  between  their  radii. 

7.  Through  any  vertex  of  a  regular  pentagon  two  diagonals  can  be 
drawn.    Find  in  degrees  the  angle  between  them. 


369-375] 


MEASUREMENT  OF  THE  CIRCLE 


251 


Section  III 


PROBLEMS 


Proposition   X 
373.    To  inscribe  a  square  in  a  given  circle. 


First,  suppose  ABCD  is  the  required  square. 

Draw  the  diagonals  AC  and  BD. 

These  are  equal  and  bisect  each  other  at  right  angles.    Prove. 

Hence  each  is  a  diameter. 

Therefore,  to  inscribe  a  square  in  a  circle,  draw  two  diame- 
ters at  right  angles  and  join  their  extremities  in  order. 

Make  the  construction  in  this  order  and  prove  that  the 
resulting  figure  is  a  square. 

374.  Theorem.  If  r  is  the  length  of  a  radius  of  a  given 
circle,  and  a  is  the  side  of  an  inscribed  square,  then  a  =  r V2. 

The  proof  follows  easily  from  the  application  of  Proposition  VII, 
Chapter  IV. 

375.  Corollary.  All  squares  inscribed  in  the  same  circle 
are  identically  equal. 

EXERCISES 

1,  Inscribe  a  regular  octagon  (polygon  of  eight  sides)  in  a  given  circle. 

2.  A  square  is  inscribed  in  a  circle  of  radius  3  inches  ;  find  the  length, 
of  each  sid^  s-hd  its  distance  from  the  centre  of  the  circle. 


252  ELEMENTARY  GEOMETRY  [Chap.  \ 

Proposition  XI 
•   376.   To  inscribe  a  regular  hexagon  in  a  given  circle. 


If  ABCDEF  is  the  required  hexagon,  and  0  the  centre  of 
the  given  circle,  each  side  of  the  hexagon  subtends  at  the 
centre  an  angle  equal  to  J-  of  four  right  angles,  or  |  of  one  right 
angle.     Therefore  A  AOB  is  equilateral,  and  AB—  OA. 

Hence,  to  construct  a  regular  inscribed  hexagon,  mark  off 
with  the  compasses  arcs  of  the  circle  whose  chords  are  each 
equal  to  a  radius,  and  join  the  points  of  division  in  order. 

377.  Theorem.  Tlie  side  of  a  regular  inscribed  hexagon  is 
equal  to  a  radius  of  the  circle. 

378.  Theorem.  If  the  alternate  vertices  of  a  regidar  inscribed 
hexagon  be  joined,  the  figure  so  formed  is  an  inscribed  equilateral 
triangle. 

379.  Theorem.  The  side  of  an  inscribed  equilateral  triangle 
is  equal  to  r  V3,  and  its  distance  from  the  centre  is  ^  r. 

380.  Definition.  The  straight  lines  joining  the  pairs  of 
opposite  vertices  in  a  polygon  of  an  even  number  of  sides  are 
called  the  principal  diagonals  of  the  polygon. 

EXERCISES 

1.  The  pairs  of  opposite  sides  of  a  regular  inscribed  hexagon  are 
parallel. 

2.  Each  principal  diagonal  of  a  regular  inscribed  hexagon  is  a  diameter. 


376-382]  MEASUREMENT  OF  THE  CIRCLE  253 

Proposition  XII 
381.   To  inscribe  a  regular  decagon  in  a  circle. 

A 


Construction.     Take  any  radius  OA,  and  divide  it  at  C  in 
extreme  and  mean  ratio,  OC  being  the  greater  segment,  so  that 

OA:OC=OC:  CA.  (Art.  275.) 

From  A  draw  a  chord  AB  equal  to  OC.     Join  OB  and  BC. 

Proof.  Since  OA:AB  =  AB:AC,  As  OAB  and  BAC  are 
similar.  (Art.  251.) 

Therefore  Z  AOB  =  Z  ABC. 

And  AB  =  BG,  since  A  5^0  must  be  isosceles. 

But  since  AB=OC,  BC=  OC, 

and  Z  COB  or  ^05  =  Z  CBO. 

That  is  Zs  ^5(7  and  CBO  are  each  equal  to  Z  ^OB. 

Therefore  Z  AOB  =  one-half  Z  05^. 

Now  Z  0^5  =  Z  OBA. 

Therefore  Z  ^ OB  is  one-fifth  of  two  right  angles,  or  one- 
tenth  of  four  right  angles. 

Therefore  AB  is  one  side  of  a  regular  inscribed  decagon. 

382.  Theorem.  If  the  alternate  vertices  of  a  regular  inscribed 
decagon  are  joined,  the  figure  so  formed  is  a  regular  inscribed 
pentagon. 


254  ELEMENTARY  GEOMETRY  [Chap.  V 

383.  Theorem.  If  from  any  point  A  of  a  circle  ttvo  chords 
AB  and  AC  are  drawn  on  the  same  side  of  A,  the  first  a  side  of  a 
regular  inscribed  decagon^  and  the  second  a  side  of  a  regular 
inscribed  hexagon,  then  the  chord  BC  is  a  side  of  a  regular 
inscribed  polygon  of  fifteen  sides. 

Suggestion.  What  fraction  of  four  right  angles  does  the  chord  AB 
subtend  at  the  centre  ?    AC?    BG  f 

Proposition  XIII 

384.  Given  the  radius  of  a  circle  and  a  side  of  an 
inscribed  regular  polygon  of  n  sides;  to  find  the  length 
of  a  side  of  an  inscribed  regular  polygon  of  2n  sides. 


Let  OA  (=  r)  be  a  radius  of  the  given  circle,  AB  (=p)  be  a 
side  of  an  inscribed  regular  polygon  of  n  sides,  and  AC  (=  g) 
be  a  side  of  an  inscribed  regular  polygon  of  2n  sides.  The 
problem  is  to  find  the  value  of  q  in  terms  of  r  and  p. 

Join  OC.     This  line  bisects  AB  at  right  angles,  at  K. 

Then  ok' =  02"-  AK'  (Art.  317 ) 

_4or^—p^ 


4 


or  OK 

whence  KO=r-^'^'l-P'=^'~^y~P'. 


383-384]  MEASUREMENT  OF  THE  CIRCLE  255 


Now  AC'  =  KC  +AK% 


or 


whence  "  g'  =  v  2  r^  —  r  V4  r^  —  p\ 

Writing  -  instead  of  r,  this  formula  becomes 
2 


^=Vf-f^^^-^' 


1  V2d2_2c«V(^'-i)'. 


If  now  we  let  Qgn  represent  the  perimeter  of  an  inscribed 
regular  polygon  of  2n  sides,  while  p  represents  the  length  of 
one  side  of  a  regular  inscribed  polygon  of  n  sides,  we  have 


Q2„=2nq=n^2d^-2d-Vd^-p\ 


EXERCISES 

1.  Does  the  above  formula  for  the  value  of  q  hold  true  when  p  =  2  r,  in 
which  case  g  is  a  side  of  an  inscribed  square  ?  Compare  the  result  with 
that  given  on  page  251. 

2.  Apply  the  formula  to  find  the  side  of  an  inscribed  regular  octagon. 

3.  In  a  circle  of  radius  r,  the  side  of  an  inscribed  cegular  dodecagon 
equals  r  V  2  —  VS. 

4.  In  a  circle  of  radius  3  feet,  what  is  the  length  of  a  side  of  an  inscribed 
regular  dodecagon  ? 

5.  The  area  of  a  regular  octagon  inscribed  in  a  circle  is  equal  to  the 
product  of  the  sides  of  the  inscribed  and  circumscribed  squares. 

6.  The  square  of  the  side  of  an  inscribed  equilateral  triangle  is  equal 
to  the  sum  of  the  squares  of  the  sides  of  the  inscribed  square  and  of  the 
inscribed  regular  hexagon. 


256  ELEMENTARY   GEOMETRY  [Chap.  V 

Proposition  XIV 

385.   To  find  approximately  the  value  of  ir. 

First,  find  the  perimeter  of  a  regular  inscribed  polygon  of  12 
sides. 

In  the  formula  of  the  preceding  problem 

let  2  n  =  12,  then  7i  =  6,  and  p  =  r  =  -  •      (Art.  377.) 


Then  Qi,  =  6\2d'-2dyjd'-  -. 


=  6  d  V2  -  V3 

=  d  X  3.105828. 

Next,  find  the  perimeter  of  a  regular  inscribed  polygon  of  24 
sides. 

If  2  n  =  24,  n  =  12,  and  p  =  ^'  =  d  X  .258819, 

§24  =  12  V2 d'-2d  Vd^ -{dx  .258819)2 

=  12  d  V2  -  2  Vr^.258819)2 

=  dx  3.132628. 
Similarly, 

§48    =t^x  3.139350 

Qgg   =dx  3.141032 

Q,^=dx  3.141452 

Q^  =dx  3.141557 

Q,es=dx  3.141584 

§1536  =  dx  3.1415904 

When  the  number  of  sides  of  the  inscribed  polygon  is  indefi- 
nitely increased,  Q2n  approaches  the  circumference  C  as  its 
limit. 

But  0=dX7r.  (Art.  362.) 


385]  MEASUREMENT  OF  THE  CIRCLE  257 

Hence  the  multipliers  of  d  in  the  above  computation  are 
approximate  values  of  tt  corresponding  to  inscribed  polygons  of 
12,  24,  48,  96  •  •  •  sides,  respectively. 

The  value  tt  =  3.1415904  is  correct  to  five  decimal  places. 

If  the  number  of  sides  of  the  polygon  weve  again  doubled,  a 
still  closer  approximation  v^ould  be  obtained. 

The  value  of  tt  has  been  computed  correctly  to  over  seven 
hundred  decimal  places.  In  practice  it  is  customary  to  use 
TT  =  3.1416. 

TT  =  -2y2-  is  an  approximate  value  correct  to  two  decimal  places, 
while  TT  =  fff  ^s  correct  to  five  decimal  places. 

A  rough  approximation  to  the  value  of  ir  can  be  obtained  by  rolling  a 
circular  piece  of  cardboard,  of  say  6  inches  radius,  along  a  straight  line 
until  it  makes  a  complete  revolution,  then  measuring  carefully  the  length 
of  the  line-segment  so  traced,  and  comparing  it  with  the  diameter  of  the 
cardboard.  A  smaller  error  will  probably  be  made  if  the  cardboard  is 
given  two  or  three  revolutions,  and  the  whole  distance  divided  by  the 
number  of  revolutions. 

A  second  method  is  to  mark  out  a  circle  on  a  piece  of  cardboard  ruled 
in  squares,  ordinary  centimetre  paper,  then  find  approximately  the  area  of 
the  circle  by  counting  the  squares  enclosed  by  it,  reckoning  each  frac- 
tional square  enclosed  by  the  circle,  a  half  square,  this  being  an  approxi- 
mate average  value.  Setting  this  area  equal  to  wr^  gives  an  approximate 
value  for  tt. 

EXERCISES 

In  these  exercises  use  ^^  as  the  value  of  ir. 

1.  The  diameter  of  a  circle  is  5  feet,  what  is  its  circumference  ? 

2.  The  radius  of  a  circle  is  1  foot  8  inches,  what  is  its  circumference  ? 

3.  A  wheel  is  twelve  feet  in  circumference,  what  is  its  diameter  ? 

4.  A  circular  field  is  1000  yards  in  circumference,  what  is  its  diameter 
and  its  area  ? 

5.  Two  fields  are  each  1600  yards  around.  One  is  circular  and  the 
other  is  square.     What  is  the  difference  in  their  areas  ? 

6.  From  a  circular  piece  of  paper  of  10  inches  radius,  a  circular  piece 
is  cut  which  has  a  radius  of  the  first  for  its  diameter.  Find  the  area  of 
the  remaining  piece. 


258  ELEMENTARY  GEOMETRY  [Chap.  V 


MISCELLANEOUS    EXERCISES 

1.  Divide  a  given  circle  into  two  arcs  such  that  any  angle  inscribed  in 
one  arc  is  three  times  an  angle  inscribed  in  the  other. 

Suggestion.  The  side  of  an  inscribed  square  is  the  chord  of  the 
required  arcs. 

2.  Divide  a  given  circle  into  two  arcs  such  that  any  angle  inscribed  in 
one  arc  is  five  times  an  angle  inscribed  in  the  other  arc. 

3.  From  a  point  without  a  circle  two  tangents  are  drawn  which,  with 
their  chord  of  contact,  form  an  equilateral  triangle  whose  side  is  18  inches. 
Find  the  diameter  of  the  circle. 

4.  In  a  regular  polygon  of  n  sides  the  straight  lines  which  join  any 
vertex  to  the  non-adjacent  vertices  divide  the  angle  at  that  vertex  into 
n  —  2  equal  parts. 

5.  An  equilateral  triangle  and  a  regular  hexagon  are  inscribed  in  a 
given  circle  ;  show  that  — 

(1)  The  area  of  the  triangle  is  half  that  of  the  hexagon ; 

(2)  The  square  on  a  side  of  the  triangle  is  three  times  the  square  on  a 
side  of  the  hexagon. 

6.  If  ABODE  is  a  regular  pentagon  and  AC^  BE  intersect  at  IT,  show 
that  — 

(1)  CH  and  EH  are  each  equal  to  a  side  of  the  pentagon  ; 

(2)  J.B  is  a  tangent  to  the  circle  circumscribed  about  the  triangle  BHG. 

7.  Show  that  the  area  of  a  regular  hexagon  inscribed  in  a  circle  is 
three-fourths  of  that  of  the  corresponding  circumscribed  hexagon. 

8.  The  area  of  a  square  circumscribed  about  a  circle  is  double  of  the 
area  of  the  inscribed  square. 

9.  If  ABGD  is  a  square  inscribed  in  a  circle  and  P  is  any  point  on  the 
circle,  show  that  the  sum  of  the  squares  on  PA,  PB,  PC,  PD  is  double 
the  square  on  the  diameter. 

10.  An  equilateral  triangle  is  inscribed  in  a  circle,  and  tangents  are 
drawn  at  its  vertices ;  prove  that  — 

(1)  The  resulting  figure  is  an  equilateral  triangle ; 

(2)  Its  area  is  four  times  that  of  the  given  triangle. 

11.  What  is  the  area  of  a  circular  ring  if  the  radii  of  the  outer  and 
inner  circles  are  132  and  120  feet,  respectively  ? 

12.  A  carriage  wheel  makes  168  revolutions  in  going  half  a  mile. 
What  is  its  height? 


386]  MEASUREMENT  OF  THE  CIRCLE  259 

13.  The  diameter  of  a  circle  is  12  feet.  Find  the  diameter  of  a  circle 
having  (1)  twice  its  circumference,  (2)  twice  its  area. 

14.  A  circular  pond  is  surrounded  by  a  gravel  walk,  such  that  the 
area  of  the  walk  equals  the  area  of  the  pond.  What  is  the  ratio  of  the 
diameter  of  the  pond  to  the  width  of  the  walk  ? 

15.  A  bicycle  wheel  is  28  inches  in  diameter.  How  many  revolutions 
will  it  make  in  going  3  miles  ? 

16.  Three  circles  are  concentric  and  are  such  that  the  area  of  the  first 
equals  the  area  between  the  first  and  the  second,  and  also  between  the 
second  and  third.  The  radius  of  the  smallest  is  10  inches.  What  are  the 
radii  of  the  other  two  ? 

17.  A  circle  is  inscribed  in  a  given  square.  What  fraction  of  the  area 
of  the  square  lies  outside  of  the  circle  ?  Is  the  fraction  changed  by 
enlarging  the  square  and  the  inscribed  circle  ? 

18.  A  square  is  inscribed  in  a  given  circle.  What  fraction  of  the  area 
of  the  circle  lies  outside  of  the  square  ? 

19.  Show  that  the  altitude  of  an  equilateral  triangle  is  to  the  radius  of 
the  circumscribed  circle  in  the  ratio  of  3  to  2. 

20.  Prove  that  the  area  of  a  circular  ring  is  equal  to  the  area  of  a 
circle  whose  diameter  equals  a  chord  of  the  outer  boundary  which  is 
tangent  to  the  inner. 

21.  Prove  that  any  equilateral  polygon  circumscribing  a  circle  is 
regular. 

22.  Each  side  of  an  inscribed  equilateral  triangle  is  parallel  to  the  tan- 
gent at  the  opposite  vertex. 

23.  Describe  a  circle  whose  area  is  equal  to  the  sum  of  the  areas  of  two 
given  circles. 

24.  Describe  a  circle  whose  circumference  is  equal  to  the  sum  of  the 
circumferences  of  two  given  circles. 

26.  Prove  that  the  radius  of  an  inscribed  regular  polygon  is  a  mean 
proportional  between  its  apothem  and  the  radius  of  a  similar  circum- 
scribed polygon. 

26.  If  squares  are  described  outwardly  on  the  sides  of  a  regular  hexa- 
gon, prove  that  the  outer  vertices  of  the  squares  are  the  vertices  of  a 
regular  dodecagon. 

27.  Find  in  degrees  the  angle  at  the  centre  of  a  circle  of  radius  2  feet, 
which  is  subtended  by  an  arc  whose  length  is  18  inches. 


260  ELEMENTARY  GEOMETRY  [Chap.  V 

SUMMARY  OF   CHAPTER  V 

1.  Definitions. 

(1)  Regular  Polygon  —  one  which  is  both  equilateral  and  equian- 

gular.    §  335. 

(2)  Centre  of  a  Regular  Polygon  — the  common  centre  of  the  in- 

scribed and  circumscribed  circles.     §  344. 

(3)  Radius  of  a  Regular  Polygon  —  the  radius  of  the  circumscribed 

circle,  i.e.  the  line  from  the  centre  to  a  vertex.     §  345. 

(4)  Apothem  of  a  Regular  Polygo7i  —  the  radius  of  the  inscribed 

circle,  i.e.  the  perpendicular  from  the  centre  on  a  side,     §  346. 

(5)  Circumference  of  a   Circle  —  the  limit  of   the  perimeter  of  a 

regular   inscribed  polygon  as  the  number  of  its  sides  is  in- 
definitely increased.     §  358. 
Or,  the  limit  of  the  perimeter  of  a  regular  circumscribed  polygon 
as  the  number  of  its  sides  is  indefinitely  increased.     §  360. 

(6)  Length  of  an  Arc  of  a  Circle  —  the  limit  of  the  sum  of  chords  in 

the  arc  as  the  number  of  such  chords  is  indefinitely  increased. 
§358. 

(7)  Area  of  a  Circle  —  the  surface  enclosed  by  the  circle,  equal  to 

the  limit  of  the  area  of  a  regular  inscribed  polygon  as  the 
number  of  its  sides  is  indefinitely  increased.     §  359. 

(8)  Similar  Arcs  of  Circles  —  arcs  which  subtend  equal  angles  at  the 

centres  of  the  circles.     §  367. 

(9)  Similar  Segments  —  those  which  have  similar  arcs.     §  368. 

(10)  Similar  Sectors  —  those  which  have  similar  arcs.     §  369. 

(11)  Principal  Diagonals  of  a  Polygon  —  the  diagonals  joining  pairs 

of  opposite  vertices  in  a  polygon  of  an  even  number  of  sides. 
§380. 

(12)  Pentagon.,  Hexagon,  Octagon,  Decagon,  Dodecagon.     See  §  336. 

2.  Postulates. 

(1)  A  circle  may  be  divided  into  any  given  number  of  equal  parts. 
(Postulate  8.)     §  337. 

3.  Problems. 

(1)  To  inscribe  a  square  in  a  given  circle.     §  373. 

(2)  To  inscribe  a  regular  hexagon  in  a  given  circle.     §  376. 

(3)  To  inscribe  an  equilateral  triangle  in  a  given  circle.     §  378. 

(4)  To  inscribe  a  regular  decagon  in  a  given  circle.     §  381. 

(5)  To  inscribe  a  regular  pentagon  in  a  given  circle.     §  382. 


Summary]         MEASUREMENT  OF  THE  CIRCLE  261 

(6)  To  inscribe  a  regular  polygon  of  fifteen  sides  in  a  given  circle. 

§383. 

(7)  To  find  the  length  of  a  side  of  an  inscribed  regular  polygon  of  2  w 

sides,  having  given  the  length  of  the  side  of  the  inscribed  poly- 
gon of  71  sides.     §  384. 

(8)  To  find  approximately  the  ratio  of  the  circumference  to  the 

diameter  of  a  circle,  i.e.  the  value  of  ir.     §  386. 
The  last  two  are  numerical  rather  than  geometrical  problems. 

4.  Theorems  on  Inscribed  and  Circumscribed  Regular  Polygons. 

(1)  If  a  circle  is  divided  into  any  number  of  equal  arcs,  (1)  the 

chords  joining  the  points  of  division  taken  in  order  form  a 
regular  inscribed  polygon,  (2)  the  tangents  to  the  circle  at 
the  points  of  division  taken  in  order  form  a  regular  circum- 
scribed polygon.     §  339. 

(2)  An  equilateral  polygon  inscribed  in  a  circle  is  also  equiangular, 

and  hence  regular.     §  342. 

(3)  A  circle  can  be  circumscribed  about  any  regular  polygon,  and 

another  circle  can  be  inscribed  in  it.     §  343. 

(4)  If  the  number  of  sides  of  a  regular  inscribed  polygon  be  doubled, 

the  perimeter  will  be  increased,  but  if  the  number  of  sides  of  a 
regular  circumscribed  polygon  be  doubled,  the  perimeter  will 
be  diminished.     §  354. 

(5)  The  area  of  a  regular  inscribed  polygon  is  increased,  and  the  area 

of  a  regular  circumscribed  polygon  is  diminished,  when  the 
number  of  sides  is  doubled.     §  355. 

(6)  The  side  of  an  inscribed  square  is  equal  to  r-\/2.     §  374. 

(7)  All  squares  inscribed  in  the  same  circle  are  identically  equal. 

§  375. 

(8)  The  side  of  a  regular  inscribed  hexagon  is  equal  to  a  radius  of 

the  circle.     §  377. 

(9)  The  side  of  an  inscribed  equilateral  triangle  is  equal  to  r\/3, 

and  its  distance  from  the  centre  is  ^  r.     §  379. 

5.  Theorems  on  the  Properties  of  Regular  Polygons. 

(1)  Any  radius  of  a  regular  polygon  bisects  the  angle  at  the  vertex. 

§  347. 

(2)  The  angle  formed  by  two  consecutive  radii  of  a  regular  polygon 

equals  four  right  angles  divided  by  the  number  of  sides  of  the 
polygon.     §  348. 

(3)  Any  two  regular  polygons  of  the  same  number  of  sides  are 

similar.     §  349. 


262  ELEMENTARY  GEOMETBY  [Chap.  ^ 

(4)  Homologous  sides  in  two  regular  polygons  of  the  same  number- 

of  sides  are  in  the  same  ratio  as  tlie  radii  of  the  circumscribed 
circles,  or  as  the  radii  of  the  inscribed  circles ;  that  is,  as  the 
radii  of  the  polygons  or  as  the  apothems  of  the  polygons.    §  350. 

(5)  The  perimeters  of  two  regular  polygons  of  the  same  number  of 

sides  are  in  the  same  ratio  as  their  radii,  or  as  their  apothems. 
§351. 

(6)  The  areas  of  two  regular  polygons  of  the  same  number  of  sides 

are  in  the  same  ratio  as  the  squares  of  their  radii,  or  as  the 
squares  of  their  apothems.     §  352. 

(7)  The  area  of  a  regular  polygon  is  equal  to  half  the  product  of  its 

perimeter  and  its  apothem.     §  353. 

6.  Theorems  on  the  Circumferences  and  Areas  of  Circles. 

(1)  The  circumference  of  a  circle  is  greater  than  the  perimeter  of  any 

regular  inscribed  polygon,  and  less  than  the  perimeter  of  any 
regular  circumscribed  polygon.     §  361. 

(2)  The  ratio  of  the  circumference  of  a  circle  to  its  diameter  is  the 

same  for  all  circles.     C  =  ird  =  2  tv.     §  362. 

(3)  The  area  of  a  circle  is  equal  to  one-half  the  product  of  its  circum- 

ference and  radius.     §  364.     A  =  ^Cr  =  irr^  =  l  tcP.     §  365. 

(4)  The  areas  of  two  circles  are  in  the  same  ratio  as  the  squares  of 

their  radii,  or  as  the  squares  of  their  diameters.     §  366. 

7.  Theorems  on  Sectors  of  a  Circle. 

(1)  The  area  of  a  sector  of  a  circle  bears  the  same  ratio  to  the  area 

of  the  circle  as  the  length  of  its  arc  bears  to  the  whole  circum- 
ference.    §  370. 

(2)  The  area  of  a  sector  of  a  circle  equals  one-half  the  product  of  its 

arc  and  radius.     S=lar.     §371. 

(3)  The  arcs  of  similar  sectors  are  in  the  same  ratio  as  the  radii, 

and  the  areas  of  similar  sectors  are  in  the  same  ratio  as  the 
squares  of  the  radii.     §  372. 


PART   II  — SOLID   GEOMETRY 


aXKc 


CHAPTER  VI 
LINES  AND  PLANES   IN  SPACE 

Section  I 

INTEESECTING  PLANES— PAKALLELS  AND  PERPENDICULARS 

386.  We  come  now  to  the  study  of  geometrical  figures  whose 
points  and  lines  do  not  all  lie  in  the  same  plane.  Such  a  figure 
is  called  a  solid  figure,  or  a  figure  of  three  dimensions. 

A  solid  figure  is  spread  out  or  extended  in  three  ways. 
A  plane  figure  is  two-dimensional  since  it  is  spread  out  in  but 
two  ways. 

A  straight  line  is  one-dimensional  since  it  has  only  length. 

387.  It  should  be  remembered  that  in  geometry  we  are  con- 
cerned only  with  forms  and  relations,  not  at  all  with  matter ; 
and  so  it  is  necessary  to  distinguish  between  a  three-dimensional 
figure,  or  a  so-called  solid  figure,  and  a  physically  solid  body. 

A  solid  figure  is  a  combination  of  points,  lines,  and  surfaces. 
It  contains  no  matter. 

Since  in  Solid  Geometry  the  diagrams  are  intended  to  represent  figures 
which  do  not  lie  wholly  in  one  plane,  they  are  drawn  as  we  say  in  per- 
spective, to  give  an  idea  of  how  the  figure  would  look  at  a  distance. 
Straight  lines  and  planes  are  supposed  to  extend  indefinitely  in  any  of 
their  directions,  though  they  must  be  represented  in  a  diagram  by  limited 
portions  of  a  line  or  a  plane.  For  the  most  part  lines  are  dotted  when 
they  are  supposed  to  be  seen  through  a  surface  which  forms  a  part  of  the 
figure. 

263 


264  ELEMENTARY  GEOMETRY  [Chap.  VI 

388.  When  we  said  (p.  3)  that  two  points  determine  a 
straight  line,  we  meant  that  through  two  points  there  can  pass 
one  straight  line,  and  only  one. 

In  just  the  same  way,  three  points  which  do  not  lie  on  the 
same  straight  line  determine  a  plane,  and  the  possibility  of 
constructing  this  plane  is  assumed. 

Postulate  9.  Through  three  points  not  in  the  same  straight 
line  there  can  he  passed  one  and  only  one  plane. 

If  then  we  name  three  points  in  any  plane,  not  lying  in  the 
same  straight  line,  they  will  be  sufficient  to  distinguish  this 
plane  from  every  other. 

Through  two  points  (or,  through  the  straight  line  joining  two 
points),  any  number  of  planes  can  be  passed. 

It  follows  immediately  that 

A  straight  line  and  a  point  not  lying  on  it  determine  a  plane. 

If  two  points  are  chosen  on  the  given  line,  these  two  and  the 
given  point  determine  the  plane.  Does  the  given  line  lie 
wholly  in  that  plane  ?     (See  definition  of  a  plane,  p.  4.) 

Two  intersecting  straight  lines  determine  a  plane. 

The  point  of  intersection  and  one  other  point  on  each  line 
determine  the  plane.  Do  each  of  the  given  lines  lie  in  that 
plane  ? 

Two  parallel  straight  lines  determiyie  a  plane. 

For,  two  parallel  straight  lines  lie  in  one  plane  by  definition 
(Art.  85),  and  this  must  be  the  plane  determined  by  one  of 
them  and  a  point  of  the  other. 

389.  Two  straight  lines  which  intersect  have  in  common  a 
single  point.  A  straight  line  and  a  curved  line,  or  two  curved 
lines,  may  have  in  common  more  than  one  point. 

A  straight  line  and  a  surface  intersect  in  one  or  more  points. 
If  the  surface  is  plane  a  straight  line  can  intersect  it  in  but 
one  point. 

Two  surfaces  intersect  in  one  or  more  lines.  These  contain 
all  points  common  to  the  two  surfaces. 


388-390]  LINES  AND  PLANES  IN  SPACE  265 

Proposition  I 
390.   The  intersection  of  two  planes  is  a  straight  line. 

p 


Let  HKL  and  PQR  be  any  two  intersecting  planes. 

It  is  required  to  prove  that  their  intersection  is  a  straight 
line,  or  in  other  words,  that  the  locus  of  points  common  to  the 
two  planes  is  a  straight  line. 

Proof.  First,  let  A  and  B  be  any  two  points  common  to  the 
planes  HKL  and  PQB. 

From  the  delinition  of  a  plane  every  point  of  the  straight 
line  AB  lies  in  the  plane  HKL.  (Art.  7.) 

For  the  same  reason,  every  point  of  the  straight  line  AB  lies 
in  the  plane  PQR. 

Therefore  the  straight  line  AB  is  common  to  the  two  planes. 

Next,  no  point  outside  of  AB  can  be  common  to  the  two 
planes,  for  then  the  two  planes  would  coincide.  (Art.  388.) 

Therefore  the  intersection  of  the  planes  HKL  and  PQR  is  a 
straight  line. 

When  no  confusion  is  likely  to  arise  a  plane  may  be  denoted  by  a  single 
letter,  as  H,  or  P,  in  the  above  diagram. 


266 


ELEMENTABT  GEOMETRY 


[Chap.  VI 


A'' 


G 


391.  Theorem.  If  two  planes  have  one  point  in  common 
they  must  have  a  second  point,  and  hence  a  straight  line,  in 
common. 

Let  the  planes  M  and  iV 
have  the  point  A  in  common. 

To  prove  that  they  must 
have  a  second  point  in 
common. 

In  the  plane  iV  choose  two 
points  B  and  C  on  the  same 
side  of  the  plane  M.  Join 
BA  and  produce  to  F,  so  that 
B  and  i^are  on  opposite  sides 
of  M.  Then  C  and  F  are  also  on  opposite  sides  of  3f,  and  the 
straight  line  CF  must  penetrate  the  plane  M  at  some  point  G 
different  from  A. .  But  since  CF  lies  wholly  in  the  plane  N,  the 
point  G  is  common  to  the  two  planes. 

By  the  preceding  theorem  the  straight  line  AG  is  common 
to  the  two  planes  and  is  the  locus  of  their  common  points. 


N 


Definitions 

392.  If  a  straight  line  and  a  given  plane  do  not  meet,  how- 
ever far  they  are  extended,  they  are  said  to  be  parallel. 

A  straight  line  not  lying  in  a  given  plane,  and  not  parallel  to 
it,  intersects  the  plane  at  one  point.  For  convenience,  this  point 
is  sometimes  called  the  foot  of  the  line. 

393.  A  straight  line  is  perpendicular  to  a  given  plane  when, 
and  only  when,  it  is  perpendicular  to  every  straight  line  in  the 
plane,  drawn  through  their  point  of  intersection.  In  that  case 
the  plane  is  also  said  to  be  perpendicular  to  the  line. 

A  straight  line  neither  parallel  nor  perpendicular  to  a  given 
plane  is  said  to  be  oblique  to  it. 

Two  planes  are  parallel  when  they  do  not  meet,  however  far 
they  may  be  extended. 


391-394]  LINES  AND  PLANES  IN  SPACE  267 


Proposition  II 

394.  If  three  planes  intersect,  two  and  two,  their  three 
lines  of  intersection  are  either  concurrent,  or  are  parallel 
two  and  two. 


-^O 


Let  L,  M,  and  N  be  the  three  given  planes,  and  let  a  be  the 
intersection  of  L  and  M,h  the  intersection  of  L  and  N,  and  c 
the  intersection  of  M  and  N. 

It  is  required  to  prove  that  the  straight  lines  a,  b,  and  c  all 
pass  through  one  point,  or  are  parallel,  two  and  two. 

Proof.  First,  suppose  the  lines  a  and  b,  which  both  lie  in 
the  plane  L,  to  intersect  at  a  point  0.  Then  because  0  is  a 
point  of  a  it  lies  on  the  plane  M,  and  because  it  is  a  point  of  b 
it  lies  on  the  plane  N. 

Therefore,  0  is  a  point  of  c,  the  intersection  of  M  and  N. 

That  is,  a,  b,  and  c  all  pass  through  one  point. 

Next,  suppose  a  and  b  are  parallel. 

Then,  if  a  and  c  are  not  parallel,  they  must  intersect,  since 
they  both  lie  in  the  plane  M;  and  it  may  be  shown  just  as 
before  that  their  common  point  lies  upon  6,  which  is  impossi- 
ble since  a  and  b  are  parallel. 

Therefore  a  and  c  must  be  parallel. 

Similarly,  b  and  c  must  be  parallel. 


268  ELEMENTARY  GEOMETRY  [Chap.  VI 


Proposition  III 

395.  The  two  planes  determined  hy  two  given  parallel 
lines  and  a  point  not  lying  in  their  plane  intersect  in  a 
line  parallel  to  each  of  the  given  lines. 


Let  a  and  h  be  the  given  parallel  lines  lying  in  the  plane  Z, 
and  let  C  be  the  given  point. 

Also,  let  M  be  the  plane  determined  by  C  and  a,  N  the  plane 
determined  by  C  and  6,  and  let  M  and  N  intersect  in  the  line  c. 

It  is  required  to  prove  that  c  is  parallel  to  both  a  and  h. 

Proof.  The  planes  X,  M,  and  N  intersect,  two  and  two,  in 
the  straight  lines  a,  b,  and  c. 

These  lines  must  therefore  be  concurrent,  or  be  parallel  two 
and  two.  (Prop.  II.) 

They  cannot  be  concurrent,  since  a  and  b  are  parallel,  by 
hypothesis. 

Therefore  c  must  be  parallel  to  both  a  and  b. 

396.  Corollary.  Tico  straight  lines  each  parallel  to  a  third 
line,  are  parallel  to  each  other. 


395-397] 


LINES  AND  PLANES  IN  SPACE 


269 


Proposition  IV 

397.  If  a  straight  line  is  perpendicular  to  each  of 
two  given  straight  lines  at  their  point  of  intersection,  it 
is  perpendicular  to  the  plane  of  those  lines. 


Let  the  straight  line  ^0  be  perpendicular  to  both  the 
straight  lines  BO  and  CO  at  their  common  point  0,  and  let 
M  be  the  plane  determined  by  BO  and  CO. 

It  is  required  to  prove  that  -40  is  perpendicular  to  the 
plane  M. 

Proof.  Suppose  OE  is  any  other  straight  line  drawn  in  the 
plane  M,  through  the  point  O.  Then  if  AO  is  perpendicular  to 
OE,  it  is  perpendicular  to  the  plane  M.  (Art.  393.) 

Draw  a  straight  line  in  the  plane  M  cutting  BO  and  CO  at 
B  and  C,  and  the  line  OE  at  E. 

Join  AB,  AE,  AC. 

Produce  ^0  to  Z>,  making  OD  equal  to  OA. 

Join  DB,  DE,  DC. 

Since  OB  is  a  perpendicular  bisector  of  the  line-segment  AD, 
BA  equals  BD.  (Art.  72.) 

Similarly  CA  equals  CD,  and  hence,  A  s  ABC  and  DBC  are 
identically  equal.  (Art.  53.) 

If  these  two  triangles  were  superposed,  DE  would  coincide 
with  AEf  and  is  therefore  equal  to  AE. 


270 


ELEMENTARY  GEOMETRY 


[Chap.  VI 


Since  E  and  0  are  each  equidistant  from  A  Siud  D,  EO  must 
be  a  perpendicular  bisector  of  AD.  (Art.  72.) 

Therefore  Z  AOE  is  a  right  angle,  and  consequently  ^0  is 
perpendicular  to  the  plane  M. 


398.   Corollary   I.     At  any  point  of  a  straight  line  07ie 
plane  can  be  constructed  perperidicular  to  that  line,  and  only  one. 


First,  two  perpendiculars,  OB  and  00, 
drawn  to  the  given  line  AD  at  the  given 
point  0,  and  the  plane  of  these  is  perpen- 
dicular to  the  given  line  (Prop.  IV). 

Next,  if  any  plane  Other  than  the  plane 
BOG  can  be  constructed  perpendicular  to 
AD  at  0,  say  the  plane  BOG',  it  must  cut 
the  line  AG  in  a  point  G'  different  from  C, 
and  the  line  OG'  must  then  be  perpendicu- 
lar to  AO,  which  is  impossible,  since  OG 
lies  in  the  same  plane  as  OG'  and  is  perpen- 
dicular to  AD  at  0. 


in  different  planes,  can  be 


399.  Corollary  II.  Tlirough  a  given  j)oint  not  on  a  given 
straight  line,  one  plane  and  only  one  can  be  constructed  perpen- 
dicular to  the  given  line. 

Let  P  be  the  given  point  and  a  the  given  straight  line.  In  the  plane  Pa 
draw  PA  perpendicular  to  a,  and  from  A  draw  another  line  AQ  perpen- 
dicular to  a.  The  plane  of  AP  and  AQ  passes  through  P  and  is  per- 
pendicular to  a  (Prop.  IV).  No  other  plane  can  pass  through  P  and  be 
perpendicular  to  a,  for  such  a  plane  must  either  intersect  a  at  J.  and  not 
contain  AQ,  or  else  it  must  intersect  a  at  a  point  different  from  A,  both 
of  which  suppositions  are  impossible.     Show  why  ? 

400.  Corollary  III.  Two  intersecting  planes  cannot  both  be 
perpendicular  to  the  same  straight  line. 


401.  Corollary  IV.  All  the  straight  lines  perpendicidar  to 
a  given  line  at  a  given  point  lie  in  one  plane  perpendicular  to  the 
given  line. 


397-402]  LINES  AND  PLANES  IN  SPACE  271 

Let  OB  and  OC  be  two  lines  perpendicular  to  the  straight  line  AD 
at  0.  Their  plane  is  perpendicular  to  AD  at  0. 
Let  OE  be  any  other  straight  line  perpendicular 
to  AD  at  0,  and  suppose,  if  possible,  that  it 
does  not  lie  in  the  plane  BOG.  Then  the  plane 
BOO  will  cut  the  plane  AOE  in  a  straight  line 
OE'  different  from  OE.  Since  OE'  lies  in  the 
plane  BOC,  ZAOE'  is  a  right  angle.  But  this 
is  impossible  since  ZAOE  is  a  right  angle. 
Therefore  OE  cannot  lie  out  of  the  plane  BOC. 

402.  Corollary  V.  If  a  plane  bisects  a  given  line-segment 
perpendicularly,  every  point  of  the  plane  is  equidistant  from 
the  extremities  of  the  line-segment. 

And  conversely :  Every  poi^it  ivhich  is  equidistant  from 
two  fixed  points  lies  in  the  plane  bisecting  perpendicularly  the 
line  Joining  them. 

EXERCISES 

1.  Through  four  points  not  lying  in  the  same  plane,  how  many  planes 
can  be  passed,  each  containing  three  of  the  points?  In  how  many  of 
these  planes  would  any  one  of  the  given  points  lie  ? 

2.  Through  five  points,  no  four  of  which  lie  in  the  same  plane,  how 
many  planes  can  be  passed,  each  containing  three  of  the  points  ? 

3.  Show  how  to  draw  through  a  given  point  a  straight  line  which  will 
intersect  two  given  lines  not  lying  in  the  same  plane.  Can  more  than  one 
such  line  be  drawn  ? 

Definition.  Two  straight  lines  so  situated  that  no  plane  can  contain 
them  both  are  said  to  be  skew  or  gauche  to  each  other. 

Hold  two  pencils  so  that  they  are  gauche  to  each  other. 

4.  Find  the  locus  of  points  in  a  plane  which  are  equidistant  from  two 
given  points  not  lying  in  the  plane. 

5.  Find  a  point  in  a  plane  equidistant  from  three  given  'points  not  lying 
in  that  plane. 

6.  A  three-legged  table  will  always  stand  firmly  on  a  level  floor,  while 
a  four-legged  table  will  not.    How  do  you  account  for  this  ? 


272 


ELEMENTARY  GEOMETRY 


[Chap.  VI 


Proposition  V 

403.   At  a  given  point  in  a  given  plane  to  erect  a 
perpendicular  to  the  plane. 


Let  L  be  the  given  plane  and  0  the  given  point  in  it. 

It  is  required  to  draw  through  0  a  straight  line  perpen- 
dicular to  the  plane  L. 

Construction.  Through  0  draw  any  two  straight  lines  OA 
and  OB  in  the  plane  L.  At  0  construct  a  plane  M  perpendicu- 
lar to  OA^  and  a  plane  N  perpendicular  to  OB.  (Art.  398.) 

These  two  planes,  M  and  N,  will  intersect  in  a  straight  line 
OP  which  is  perpendicular  to  L. 

Proof.  Since  OP  lies  in  the  plane  M  and  passes  through  0, 
it  is  perpendicular  to  OA.  (Art.  393.) 

And  since  it  lies  in  the  plane  N,  it  is  perpendicular  to  OB. 

Therefore  OP  is  perpendicular  to  the  plane  L.      (Prop.  IV.) 

404.  Theorem.  At  a  point  in  a  plane,  hut  one  straight  line 
can  he  drawn  perpendicular  to  the  plane. 

If  there  were  a  second  straight 
line  OQ  perpendicular  to  the  plane 
X,  the  plane  of  OP  and  0^  would 
intersect  Z  in  a  straight  line  OR^ 
to  which  OP  and  OQ  are  both  per- 
pendicular.    But  this  is  impossible. 


.^'R 


403-406]  LINES  AND  PLANES  IN  SPACE 


273 


Proposition  VI 

405.   Two  straight  lines  perpendicular  to  the  same 
plane  are  parallel. 

a  h 


A-^--": 


Let  a  and  h  be  straight  lines  perpendicular  to  the  plane 
Lj  at  the  points  A  and  B,  respectively. 

It  is  required  to  prove  that  a  and  h  are  parallel. 

Proof.     First,  to  show  that  a  and  h  are  in  the  same  plane. 

Join  AB,  and  in  the  plane  L  draw  BQ  perpendicular  to  AB. 

In  a  choose  any  point  P,  and  make  BQ  equal  to  AP. 

Join  PB,  PQ,  and  AQ. 

A  PAB  is  identically  equal  to  A  QBA.     Why  ? 

Therefore  BP=AQ. 

Also,  A  PBQ  is  identically  equal  to  A  QAP.     Why  ? 

Therefore  Z.PBQ  is  a  right  angle. 

Hence  b,  BP,  and  BA  must  lie  in  one  plane.  (Art.  401.) 

But  since  a  meets  both  BP  and  BA,  it  also  must  lie  in  that 
plane. 

Next,  the  interior  angles  which  a  and  b  make  with  the 
transversal  AB  are  both  right  angles. 

Therefore  a  and  b  are  parallel.  (Art.  90.) 

406.   Corollary  I.    If  from  the  foot  of  a  given  perpendicular 

to  a  plane,  a  straight  line  is  drawn  at  right  angles  to  any  line 

in  the  plane,  any  line  through  their  intersection,  ivhich  meets  the 

given  perpendicular,  is  at  right  angles  to  the  line  of  the  plane. 

That  is,  BP  drawn  to  meet  a  is  perpendicular  to  BQ. 


274 


ELEMENTABY  GEOMETRY 


[Chap.  VI 


407.  Corollary   II.    If  one  of  two  parallel  lines  is  per- 
pendicular to  a  plane,  the  other  is  also. 

For,  if  not,  at  its  point  of  intersection  with  the  plane  erect 
a  perpendicular.  This  lies  in  a  plane  with  the  first  perpen- 
dicular and  is  parallel  to  it. 

Proposition  VII 

408.  From  a  given  point  without  a  given  plane  to 
draw  a  perpendicular  to  the  plane. 


Let  L  be  the  given  plane,  and  P  the  given  point  without  it. 

It  is  required  to  draw  from  P  a  straight  line  perpendicular 
to  the  plane  L. 

Construction.  In  L  take  any  straight  line  a,  and  through 
P  construct  the  plane  M  perpendicular  to  a.  (Art.  399.) 

This  plane  will  intersect  L  along  some  line  AE. 
From  P  in  the  plane  M  draw  PB  perpendicular  to  AE. 
PB  is  the  required  line. 

Proof.  From  B  in  the  plane  L  draw  the  straight  line  h 
parallel  to  a. 

Then  h  is  perpendicular  to  the  plane  M  (Art.  407),  and 
hence  perpendicular  to  PB  which  lies  in  M. 

But  PB  was  drawn  perpendicular  to  AE. 

Therefore  PB,  being  perpendicular  to  both  AE  and  h,  is 
perpendicular  to  their  plane,  namely,  the  plane  L. 


407-411]  LINES  AND  PLANES  IN  SPACE  275 

409.  Theorem.  From  a  point  without  a  plane  only  one 
perpendicular  to  the  plane  can  he  drawn. 

For  if  there  are  two  perpendicu- 
lars from  P,  their  plane  will  intersect 
the  given  plane  in  a  straight  line  per- 
pendicular to  them  both,  which  is 
impossible. 

410.  Corollary.  Of  all  straight 
lines  which  can  he  drawn  from  a 
point  to  a  plane,  the  perpendicular  is  the  shortest. 

411.  Definition.     The  distance  from  a  point  to  a  plane  is 

the  length  of  the  perpendicular  from  the  point  to  the  plane. 

EXERCISES 

1.  AB  and  CD  are  two  straight  lines  of  which  AB  lies  in  a  plane 
perpendicular  to  CD.  Prove  that  the  perpendiculars  to  AB  from  the 
different  points  of  CD  all  pass  through  one  point. 

Suggestion.  Compare  this  exercise  with  the  theorem  stated  in  Article 
406.     What  is  the  relation  between  them  ? 

2.  If  D  is  any  point  on  the  straight  line  drawn  through  the  ortho- 
centre  of  a  triangle  ABC,  perpendicular  to  the  plane  of  the  triangle,  then 
DA  is  at  right  angles  to  the  straight  line  drawn  through  A  parallel 
to  BC. 

3.  Find  the  locus  of  points  in  space  equidistant  from  the  points  of  a 
circle. 

4.  Equal  oblique  lines  drawn  from  a  point  to  a  plane  make  equal 
angles  with  the  perpendicular  from  the  point,  and  meet  the  plane  at 
equal  distances  from  the  foot  of  the  perpendicular.  Of  two  unequal 
lines  so  drawn,  which  makes  the  greater  angle  with  the  perpendicular  ? 

5.  If  two  straight  lines  are  each  parallel  to  a  third  line,  they  are 
parallel  to  each  other.     Prove.     (See  Article  396.) 

Suggestion,  A  plane  perpendicular  to  the  third  line  is  perpendicular 
to  each  of  the  others. 

6.  If  two  straight  lines  are  not  in  the  same  plane,  no  two  straight  lines 
joining  points  of  the  one  to  points  of  the  other  can  be  parallel. 


276 


ELEMENTARY  GEOMETRY  [Chap.  VI 


Proposition  VIII 

412.  Two  planes  perpendicular  to  the  same  straight 
line  are  parallel. 


'M 


7 


7 


Let  L  and  M  be  two  planes,  each  perpendicular  to  the 
straight  line  a,  at  the  points  P  and  Q,  respectively. 

It  is  required  to  prove  that  these  planes  are  parallel. 

Proof.  If  L  and  M  are  not  parallel  they  must  intersect 
in  some  straight  line.  Let  E  be  any  point  of  their  inter- 
section. 

Then  through  E  there  would  pass  two  planes  perpendicular 
to  a,  which  is  impossible.  (Art.  399.) 

Therefore  L  and  M  are  parallel. 


EXERCISES 

1.  If  two  planes  are  perpendicular  to  the  same  straight  line,  any  plane 
containing  that  line  intersects  them  in  parallel  lines. 

2.  Find  the  locus  of  points  equidistant  from  two  given  parallel  planes. 

3.  Prove  that  through  a  given  point  outside  a  given  plane,  any  required 
number  of  straight  lines  can  be  drawn  parallel  to  the  given  plane. 


412-414] 


LINES  AND  PLANES  IN  SPACE 


277 


Proposition  IX 

413.  If  two  straight  lines  are  parallel,  any  plane  con- 
taining one  of  them,  and  not  the  other,  is  parallel  to 
the  other. 

N 


Let  a  and  6  be  two  parallel  straight  lines,  and  let  the  plane 
M  contain  the  line  6,  but  not  a. 

It  is  required  to  prove  that  the  plane  M  is  parallel  to  the 
line  a. 

Proof.  Since  a  and  h  are  parallel,  they  lie  in  one  plane  by- 
definition,  say  the  plane  N. 

All  the  points  common  to  M  and  N  lie  on  the  straight 
line  h.  (Prop.  I.) 

Therefore,  if  a  intersects  M  at  any  point,  it  must  be  at  some 
point  of  h. 

But  a  is  parallel  to  h,  by  hypothesis,  and  therefore,  can  have 
no  point  in  common  with  6,  or  with  the  plane  M.  Therefore  M 
is  parallel  to  a. 

414.  Corollary  I.  Through  either  of  two  given  straight 
lines  not  lying  in  the  same  plane,  one  plane  can  be  passed  parallel 
to  the  other  line.  a 

If  a  and  b  are  the  given  straight 
lines,  from  any  point  P  of  6  draw 
PQ  parallel  to  a.  Then  the  plane 
of  b  and  PQ  is  parallel  to  a,  and 
passes  through  b. 

Can  more  than  one  plane  be  so 
passed  through  b  ? 


278  ELEMENTAUY  GEOMETRY  [Chap.  VI 

415.  Corollary  II.  Through  a  given  point,  a  plane  can  he 
passed  parallel  to  any  tivo  given  straight  lines  in  space. 

Suggestion.  Through  the  given  point  draw  straight  lines  parallel  to 
the  given  straight  lines.  ' 

Proposition  X 

416.  If  a  straight  line  is  parallel  to  a  given  plane,  it 
is  parallel  to  the  intersection  of  any  plane  through  it, 
with  tJie  given  plane. 

The  proof  is  left  to  the  student  with  the  following  queries : 

(1)  Are  the  two  straight  lines  under  consideration  in  the 
same  plane  ? 

(2)  Will  they  meet  if  produced  ? 

417.  Corollary.  If  a  straight  line  is  parallel  to  a  given 
plane,  a  line  drawn  from  any  point  m  the  plane  parallel  to  the 
given  line  lies  in  the  given  plane. 

Suggestion.     The  chosen  point  and  the  given  line  determine  a  plane. 

EXERCISES 

1.  If  from  any  point  perpendiculars  are  dravsrn  to  two  parallel  planes, 
these  perpendiculars  must  coincide, 

2.  Prove  that  if  one  of  two  parallel  planes  is  perpendicular  to  a  given 
straight  line,  the  other  must  be. 

3.  Prove  that  if  through  each  of  two  parallel  straight  lines,  a  plane  is 
passed,  these  two  planes  will  intersect,  if  at  all,  in  a  straight  line  parallel 
to  both  the  given  lines. 

4.  If  a  and  6  are  two  straight  lines  which  do  not  lie  in  the  same  plane, 
and  c  and  d^  two  straight  lines  meeting  them,  prove  that  either  c  and  d 
intersect  on  a,  or  on  &,  or  else  have  no  common  plane. 

5.  If  two  straight  lines  are  parallel,  any  straight  line  joinhig  a  point  in 
one  of  them  to  a  point  in  the  other  lies  in  the  plane  determined  by  the 
two  parallel  lines. 


415-418] 


LINES  AND  PLANES  IN  SPACE 


279 


Proposition  XI 

418.  If  two  intersecting  straight  lines  are  each  parallel 
to  a  given  plane,  the  plane  determined  by  these  lines  is 
also  parallel  to  the  given  plane. 


A 

^7 

/' 

k  / 

Let  a  and  h  be  two  straight  lines,  intersecting  at  P,  which 
are  each  parallel  to  the  plane  L. 

It  is  required  to  prove  that  the  plane  M,  determined  by  a  and 
h,  is  also  parallel  to  L. 

Proof.     From  P  draw  a  perpendicular  to  the  plane  L,  meeting 
it  at  Q.  (Prop.  VII.) 

From  Q  draw  a'  and  6'  parallel  to  a  and  h,  respectively. 

These  will  lie  in  the  plane  L  (Prop.  X,  Cor.)  and  hence  be 
perpendicular  to  PQ. 

The  straight  lines  a  and  h,  being  parallel  to  a'  and  6'  re- 
spectively, are  also  perpendicular  to  PQ.  (Art.  97.) 

Hence  the  plane  M  is  perpendicular  to  PQ.  (Prop.  IV.) 

But  PQ  was  drawn  perpendicular  to  L. 

Therefore  the  planes  L  and  M  are  parallel  to  each  other. 

(Prop.  VIII.) 
EXERCISES 

1.  If  three  straight  lines  intersect  at  one  point,  and  all  meet  a  fourth 
straight  line  not  through  that  point,  the  first  three  must  lie  in  one  plane. 

2.  If  three  straight  lines  are  such  that  each  intersects  the  other  two, 
they  must  all  lie  in  one  plane,  or  must  all  pass  through  one  point. 


280  ELEMENTARY  GEOMETRY  [Chap.  VI 


Proposition  XII 

419.   Two  parallel  planes  are  intersected  hy  any  third 
plane  in  parallel  lines. 


Let  L  and  Jf  be  any  two  parallel  planes,  and  let  JVbe  a  third 
plane  intersecting  L  in  the  straight  line  AB,  and  M  in  the 
straight  line  PQ. 

It  is  required  to  prove  that  AB  and  PQ  are  parallel. 

Proof.     First,  AB  and  PQ  lie  in  the  same  plane  N. 
Next,  AB  and  PQ  cannot  meet  since  the  planes  L  and  M 
have  no  point  in  common  however  far  they  are  extended. 
Therefore  AB  and  PQ  are  parallel. 

420.  Corollary  I.  Parallel  line-segments  terminated  hy 
parallel  planes  are  equal. 

Pass  a  plane  through  the  two  parallel  line-segraents.  The  two 
given  line-segments  and  the  intersections  of  this  plane  with 
the  parallel  planes  form  a  parallelogram. 


419-423]  LINES  AND  PLANES  IN  SPACE  281 

421.  Corollary   II.      Two  parallel  planes  are  everywhere 
equidistant. 

422.  Corollary  III.     A  straight  line  perpendicular  to  one 
of  two  parallel  ])lanes  is  also  perpendicular  to  the  other. 

Pass  two  planes  through  the  given  line  and  study  their  inter- 
sections with  the  given  planes. 

423.  Corollary  IV.     Tlirough  any  point  one  plane  can  he 
passed  parallel  to  a  given  plane,  and  only  one. 

Let  P  be  the  given  point  and  L  the  given 
plane. 

From  P  draw  PQ  perpendicular  to  L. 

At  P  construct  a  plane  M  perpendicular 
to  PQ. 

Then  Jf  is  parallel  to  L.      (Prop.  VIII.) 

If  there  is  a  second  plane  through  P 
parallel  to  L,  PQ  must  also  be  perpen- 
dicular to  it  (Cor.  Ill) .    This  is  impossible. 

(Art.  400.) 


EXERCISES 

1.  Three  planes  in  general  intersect  at  one  point.  Can  three  planes 
have  a  straight  line  in  common  without  coinciding  ? 

2.  There  is  one  rectilinear  figure  which  is  necessarily  plane.  What 
is  it? 

3.  Four  planes  are  such  that  no  three  of  them  pass  through  the  same 
straight  line  and  no  four  of  them  contain  the  same  point.  In  how  many 
straight  lines  do  they  intersect,  two  and  two  ?  In  how  many  points  do 
they  intersect,  three  and  three  ?  How  many  lines  of  intersection  through 
each  point  ?    How  many  points  of  intersection  on  each  line  ? 

4.  A  and  B  are  two  points  on  the  same  side  of  a  plane,  P  a  point  in 
the  plane.  Determine  P  so  that  the  sum  of  the  line-segments  AP  and 
PB  shall  be  the  least  possible. 

5.  If  four  planes  are  given,  and  the  common  line  of  the  first  two  inter- 
sects the  common  line  of  the  last  two,  the  four  planes  meet  in  one  point. 


/  '  / 

/  ''  / 

282 


ELEMENTARY  GEOMETRY 


[Chap.  VI 


Proposition  XIII 

424.  If  two  intersecting  straight  lines  lying  in  one 
plane  are  parallel,  respectively,  to  two  intersecting 
straight  lines  lying  in  another  plane,  the  two  planes 
must  be  parallel,  and  the  angles  formed  hy  the  lines 
are  equal. 


Let  the  straight  lines  AB  and  AC,  lying  in  the  plane  L,  be 
parallel,  respectively,  to  the  straight  lines  A'B'  and  AO,  lying 
in  the  plane  Jf,  and  let  them  extend  in  the  same  way  from  the 
vertex. 

It  is  required  to  prove  that  the  planes  L  and  M  are  parallel, 
and  that  the  angle  BAG  equals  the  angle  B'A'C 

Proof.  First,  the  lines  AB  and  AC  are  both  parallel  to  the 
plane  M.  (Prop.  IX.) 

Therefore  the  plane  L  is  parallel  to  the  plane  M.    (Prop.  XI.) 

Next,  from  AB  and  A'B'  cufc  oft'  equal  segments  AD  and  A'D'. 

Also  from  AC  and  A'C  cut  off  equal  segments  AE  and  A'E'. 

Join  AA',  DD',  EE',  also  DE  and  D'E'. 

AA'D'D  is  a  parallelogram.     Why  ? 

Therefore  DD'  is  equal  and  parallel  to  AA\ 

AA'E'E  is  also  a  parallelogram.     Why  ? 

Therefore  EE'  is  equal  and  parallel  to  AA'. 

Hence  EE'  is  equal  and  parallel  to  DD'.  (Art.  396.) 

Therefore  EE'D'D  is  a  parallelogram,  and  ED  is  equal  to 
E'D'. 

ThereioTe  A  DAE  is  identically  equal  to  AD' A'E'  (Art.  53), 
and  Z  DAE  is  equal  to  Z  D'A'E'. 


424-425] 


LINES  AND  PLANES  IN  SPACE 


283 


Proposition  XIV 

425.  If  two  straight  lines  are  eut  by  three  parallel 
planes,  the  corresponding  segments  are  proportional. 


Let  AB  and  CD  be  two  straight  lines  which  are  intersected 
by  the  parallel  planes  L,  M,  N,  in  the  points  A,  E,  B  and 
C,  F,  D,  respectively. 

It  is  required  to  prove  that  AE :  EB  =  CF  :  FD. 

Proof.  Draw  the  straight  line  AD  cutting  the  plane  M  at 
the  point  K. 

Join  EK  2ind  KF,  also  AC  and  BD. 

Now,  the  plane  BAD  must  intersect  M  and  N  in  parallel 
lines.  (Prop.  XII.) 

Therefore  EK  is  parallel  to  BD. 

Hence  AE:EB  =  AK:  KD.  (Art.  242.) 

Similarly,  AK :  KD  =  CF :  FD. 

Therefore  AE:EB  =  CF:  FD. 


EXERCISES 

1.  If  n  parallel  planes  intercept  equal  segments  on  one  straight  line, 
they  will  intercept  equal  segments  on  any  other  straight  line. 


284 


ELEMENTARY  GEOMETRY 


[Chap.  Vi 


Section  II 

DIHEDRAL  ANGLES 

426.  When  two  planes  intersect  they  are  said  to  form  a 
dihedral  angle. 

The  line  of  intersection  of  the  planes  is  called  the  edge  of 
the  dihedral  angle,  and  the  planes  are  the  boundaries  or  faces 
of  the  angle. 

By  rotating  one  of  the  planes  about  their  line  of  intersection, 
the  other  being  held  stationary,  the  dihedral  angle  is  increased 
or    diminished,  according    as    the 
plane  is  rotated  in  the  one  way, 
or  in  the  other. 

A  dihedral  angle  may  be  desig- 
nated by  mentioning  two  points  in 
its  edge  and  one  other  point  in 
each  boundary. 

Thus  the  dihedral  angle  in  the  diagram  may  be  designated 
by  P-AB-Q. 

When  no  confusion  can  arise  it  will  be  sufficient  in  speaking 
of  a  dihedral  angle  to  name  its  edge. 

427.  Two  planes  which  extend  both  ways  from  their  line 
of  intersection  form,  in  all, 
four  dihedral  angles  having 
the  same  edge,  those  being 
adjacent  angles  which  have 
one  boundary  in  common, 
and  those  being  vertical 
angles  in  which  the  boun- 
daries of  one  are  the  ex- 
tensions of  the  boundaries 
of  the  other. 

Thus  in  the  diagram 
P-AB-Q  and  P-AB-R 
are  adjacent  angles,  P-AB-Q  and  R-AB-S  are  vertical  angles. 


426-431]  LI^ES  AND  PLANES  IN  SPACE  285 

428.  Two  dihedral  angles  are  equal  when  their  boundaries 
can  be  made  to  coincide. 

429.  Definition.  If  one  plane  meets  another  so  as  to  make 
the  adjacent  dihedral  angles  equal,  the  first  plane  is  said  to  be 
perpendicular  to  the  second,  and  the  angles  so  formed  are  called 
right  dihedral  angles. 

430.  If  we  choose  a  point  in  the  edge  of  a  dihedral  angle, 
and  from  it  draw  two  straight  lines  at  random,  one  in  each 
boundary  of  the  dihedral  angle,  the  plane  angle  formed  by 
these  lines  may  have  any  magnitude  from  zero  to  two  right 
angles,  the  dihedral  angle  remaining  unaltered. 

How  must  the  two  lines  be  drawn  in  ^. 

order  that  the  angle  between  them  shall  y^^^m. 

be  (1)  zero  ?  (2)  two  right  angles  ?  j^^^^^^W^ 

Definition.     If  from  any  point      y[y__     /^ /^    _^ 
of  the  edge  of  a  dihedral  angle,  two     ^^- — / / — ^^^^ 
straight  lines  are  drawn  perpen-         \/X  ^^m 

dicular  to  that  edge,  one  in  each        ^\/  ^       %^ 

boundary  of  the  angle,  the  plane         ^ 

angle  formed  by  these  lines  is  called  the  inclination  of  the 
planes  forming  the  dihedral  angle;  or  it  is  called  the  plane 
angle  of  the  dihedral  angle. 

Thus  in  the  diagram  Z  PQR  is  the  inclination  of  the  planes 
P  and  i?,  if  PQ  and  J?Q  are  both  perpendicular  to  the  edge  AB. 

431.  That  the  inclination  of  the  planes  is  the  same  no 
matter  from  what  point  of  the  edge  the  perpendiculars  are 
drawn  follows  from  Proposition  XIII.  For  since  QP  and  Q'P 
lie  in  the  same  plane  and  are  both  perpendicular  to  AB^  they 
are  parallel  (Art.  91).  Similarly  QP  and  Q'P'  are  parallel. 
Therefore  /.PQR  equals  ZP'Q'P' wherever  on  the  edge  Q 
may  be  chosen. 

In  other  words. 

All  plane  angles  of  the  same  dihedral  angle  are  equal. 


286 


ELEMENTARY  GEOMETRY 


[Chap.  VI 


432.  From  the  definition  of  the  plane  angle  of  a  dihedral 
angle  it  follows  that, 

The  lines  of  intersection  with  the  boundaries  of  a  dihedral 
angle,  of  any  plane  perpendicular  to  the  edge  of  thai  angle,  form 
the  plane  angle  of  the  dihedral  angle. 


Proposition  XV 

433.   Two  dihedral  angles  are  equal  if  their  -plane 
angles  are  equal. 


Let  P-AB-Q  and  P'-A'B'-Q'  be  two  dihedral  angles  whose 
plane  angles,  JfOiV  and  M'O'N',  are  equal. 

It  is  required  to  prove  that  the  dihedral  angles  are  equal. 

Proof.  Place  the  one  figure  upon  the  other  so  that  Z  M'O'N' 
coincides  with  its  equal  Z  MON. 

The  edge  AB'  must  then  coincide  with  the  edge  AB,  since 
both  are  perpendicular  to  the  plane  MON  at  the  point  0. 

(Art.  404.) 

The  plane  P'  must  coincide  with  the  plane  P,  since  two 
lines  determining  F,  viz.',  A'B'  and  O'M',  coincide  with  two 
lines  determining  P,  viz.,  AB  and  OM. 

Similarly  the  plane  Q'  must  coincide  with  the  plane  Q. 
Therefore  the  tw^o  dihedral  angles,  having  been  made  to  coin- 
cide, are  equal. 

EXERCISE 

1.   State  and  prove  the  converse  of  the  above  theorem. 


432-435]  LINES  AND  PLANES  IN  SPACE  287 

Proposition  XVI 

434.  Two  dihedral  angles  are  in  the  same  ratio  as 
their  plane  angles. 

Outline  of  proof. 

Case  I.     When  the  plane  angles  are  commensurable. 

1.  Choose  a  common  measure  for  the  plane  angles,  and 
divide  them  both  into  equal  parts.  The  numbers  of  parts  will 
give  the  ratio  of  the  plane  angles. 

2.  Through  the  edges  of  the  two  dihedral  angles  and  the 
lines  of  division  of  the  plane  angles  pass  planes.  These  will 
divide  the  dihedral  angles  into  equal  parts,  the  same  in  number 
as  the  equal  parts  of  the  plane  angles. 

3.  The  ratio  of  the  dihedral  angles  will  be  given  by  the 
numbers  of  parts  into  which  they  are  divided.  This  ratio  will 
be  the  same  as  the  ratio  of  the  plane  angles. 

Case  II.     When  the  plane  angles  are  incommensurable. 

1.  Choose  a  measure  of  one  of  the  plane  angles ;  with  it  as 
unit  divide  that  angle  into  equal  parts  and  apply  the  unit  as 
often  as  possible  to  the  other  plane  angle. 

2.  Proceed  with  the  proof  as  in  Case  II,  Prop.  X,  Chap. 
Ill,  applying  the  Principle  of  Limits.  Give  the  full  demon- 
stration. 

435.  Note.  Since  dihedral  angles  are  in  the  same  ratio  as 
their  plane  angles,  the  measure  of  the  plane  angle  gives  the 
measure  of  the  dihedral  angle.  Or  we  may  say  that  to  measure 
a  dihedral  angle,  we  measure  its  plane  angle.  Thus  we  have 
a  dihedral  angle  of  90°,  or  30°,  or  10°,  when  its  plane  angle  is 
90°,  or  30°,  or  10°. 

A  right  dihedral  angle  has  its  plane  angle  equal  to  a  right 
angle;  and  two  planes  are  thus  perpendicular  to  each  other, 
when  the  plane  angle  of  any  one  of  the  dihedral  angles  formed 
by  them,  is  a  right  angle. 


288 


ELEMENTARY  GEOMETRY 


[Chap.  VI 


Proposition  XVII 

436.  If  a  straight  line  is  perpendicular  to  a  given 
plane,  every  plane  containing  that  line  is  perpendicular 
to  tJie  given  plane. 


Let  AB  be  a  straight  line  perpendicular  to  the  given  plane 
L,  at  the  point  A,  and  let  M  be  any  plane  containing  AB. 

It  is  required  to  prove  that  the  plane  M  is  perpendicular  to 
the  plane  L. 

Proof.  From  A  draw  the  straight  line  ^C  in  the  plane  L, 
perpendicular  to  AE,  the  intersection  of  L  and  M. 

Z  BAG  is  the  plane  angle  of  the  dihedral  angle  formed  by 
L  and  M.     Why  ? 

Z  BAC  is  a  right  angle.     Why  ? 

Therefore  the  plane  M  i^  perpendicular  to  the  plane  L. 

437.  Corollary  I.  If  a  plane  is  perpendicular  to  any 
straight  line  of  another  plane,  it  is  perpendicular  to  that  other 
plane. 

438.  Corollary  II.  Any  plane  perpendicular  to  the  edge 
of  a  dihedral  angle  is  perpendicular  to  each  of  its  faces. 

EXERCISES 

1.  Through  a  given  straight  line  pass  a  plane  perpendicular  to  a  given 
plane. 


436-440] 


LINES  AND  PLANES  IN  SPACE 


289 


Proposition  XVIII 

439.  If  two  plaiies  are  perpendicular  to  each  other,  a 
straight  line  drawn  in  one  of  them,  perpendicular  to 
tl%eir  intersection,  is  perpendicular  to  the  other. 


Let  L  and  M  be  two  planes  perpendicular  to  each  other,  and 
let  PQ  be  drawn  in  the  plane  L  perpendicular  to  their  inter- 
section AB. 

It  is  required  to  prove  that  PQ  is  perpendicular  to  the  plane  M. 

Proof.    From  P  draw  PR  in  the  plane  M  perpendicular  to  PB. 

Z  PPQ  is  the  plane  angle  of  the  dihedral  angle  formed 
by  the  planes  L  and  M,  and  is  therefore  a  right  angle  by 
hypothesis. 

Since  PQ  is  at  right  angles  to  both  PB  and  PP  it  is  perpen- 
dicular to  the  plane  M.  (Prop.  IV.) 


440.  Corollary  I.  If  two  planes  are  perpendicular  to  each 
other,  a  straight  line  drawn  from  any  point  of  their  intersection, 
perpendicular  to  one  plane,  must  lie  in  the  other. 

For  if  the  line  PQ  is  drawn  in  L  perpendicular  to  the  inter- 
section of  L  and  M^  it  is  perpendicular  to  31.  Hence  no 
other  line  drawn  from  P  can  be  perpendicular  to  31  (Art.  404) . 
That  is,  the  perpendicular  to  iHf,  drawn  from  P,  must  lie 
in  L. 
u 


290 


ELEMENTARY  GEOMETRY 


[Chap.  VI 


441.  Corollary  II.  Every  plane  through  a  point,  perpen- 
dicular to  a  given  plane,  contains  the  line  from  the  point,  perpen- 
dicular to  the  giuen  plane. 

442.  Corollary  III.  If  tivo  planes  are  perpendicular  to  each 
other,  a  straight  line  drawn  from  any  p)oint  in  one  of  them,  per- 
pe7idicular  to  the  other,  must  lie  in  the  first  plane. 

For  the  line  QP,  drawn  from  Q  perpendicular  to  the  intersec- 
tion of  L  and  M,  lies  in  L,  and  is  perpendicular  to  M. 
Hence,  no  other  line  drawn  from  Q  can  be  perpendicular 
to  M.  (Art. 409.) 

That  is,  the  perpendicular  to  Jf,  drawn  from  any  point  of  X, 
must  lie  in  L. 

443.  Corollary  IV.  If  two  intersecting  planes  are  each 
perpendicular  to  a  third  plane,  their  intersection  is  also  perpen- 
dicular to  that  plane. 

If  M  and  JSf  are  each  perpendicu- 
lar to  i,  and  0  is  the  point  in  which 
L  intersects  the  common  line  of  M 
and  N,  the  perpendicular  to  L  drawn 
from  O  must  lie  in  both  M  and  N 
(Cor.  I).  Hence  it  must  be  their 
intersection.  That  is,  the  intersec- 
tion of  M  and  N  is  perpendicular 
to  L. 

EXERCISES 

1.  Construct  the  plane  which  bisects  a  given  dihedral  angle. 

2.  Through  a  given  point  pass  a  plane  perpendicular  to  each  of  two 
given  planes. 

3.  Perpendiculars  to  a  plane  are  drawn  from  the  points  of  a  straight 
line  not  lying  in  the  plane.  Show  that  the  intersections  of  these  perpen- 
diculars with  the  plane  are  all  in  one  straight  line. 

4.  Through  a  point  without  a  straight  line  any  number  of  planes  can 
pass  parallel  to  that  line. 

5.  Find  the  locus  of  points  equidistant  from  two  given  parallel  planes 
and  at  the  same  time  equidistant  from  two  given  points. 


441-444] 


LINES  AND  PLANES  IN  SPACE 


291 


Proposition  XIX 

444.  The  locus  of  points  equidistant  from  the 
boundaries  of  a  dihedral  angle  is  the  plane  bisecting 
that  angle. 


Let  the  planes  L  and  M  form  a  dihedral  angle  whose  edge  is 
AB,  and  let  the  plane  N  bisect  this  angle. 

It  is  required  to  prove  that  every  point  of  N  is  equidistant 
from  L  and  M,  and  that  no  point  outside  of  N  is  equidistant 
from  L  and  M. 

Proof.  First,  in  N  choose  any  point  P,  and  from  it  draw  PQ 
and  PR  perpendicular  to  L  and  M,  respectively.     (Prop.  VII.) 

The  plane  of  PQ  and  PR  is  perpendicular  to  both  L  and  M, 
and  hence  is  perpendicular  to  the  edge  AB.  (Art.  443.) 

Let  this  plane  intersect  AB  at  S,  and  join  SP,  SQ,  SR. 

Then  Zs  QSR,  QSP,  and  PSR  are  the  plane  angles  of  the 
dihedral  angles  L-AB-M,  L-AB-N,  and  N-AB-M,  re- 
spectively. (Art.  432.) 

Therefore  Z  QSP  =  Z  PSR  by  hypothesis. 

Also  Zs  PQS  and  PRS  are  right  angles.     Why  ? 

Therefore  As  PQS  and  PRS  are  identically  equal,  and  PQ 
equals  PR.  (Art.  106.) 

Therefore  P,  any  point  of  N,  is  equidistant  from  the  planes 
L  and  M. 

Next,  let  P'  be  any  point  outside  of  the  plane  N,  and  from  it 


292  ELEMENTARY  GEOMETRY  [Chap.  VI 

draw  perpendiculars  P'Q  and  FR^  to  the  planes  L  and  M, 
respectively.     Let  P'Q  intersect  the  plane  N  at  the  point  P. 

Join  FR.     FR'  is  less  than  FR.     Why  ? 

But  FR  is  less  than  FP  +  PR,  i.e.  less  than  FQ.     Why  ? 

Therefore  P'i?'  is  less  than  FQ. 

Hence  P'  is  not  equidistant  from  L  and  M. 

Therefore  the  locus  •  •  • 

Definitions 

445.  The  projection  of  a  point  upon  a  plane  is  the  foot  of  the 
perpendicular  drawn  from  the  point  to  the  plane. 

The  projection  of  a  line  upon  a  plane  is  the  locus  of  the 
projections  of  its  points  upon  the  plane. 

446.  Theorem.  The  projection  of  a  straight  line  upon  a 
plane  must  he  a  straight  line. 

For,  if  A  be  any  point  of  the  given  line,  and  P  its  projection 
upon  the  given  plane,  AP  and  the  given  line  determine  a 
plane  perpendicular  to  the  given  plane,  in  which  lie  the  per- 
pendiculars from  all  the  points  of  the  given  line.   (Art.  442.) 

The  projection  of  a  curved  line  upon  a  plane  is  in  general 
another  curved  line.     Might  it  in  any  case  be  a  straight  line  ? 

447.  The  angle  which  a  straight  line  makes  with  its  projec- 
tion on  a  plane  is  called  the  inclination  of  the  line  to  the  plane. 

Whenever  we  speak  of  '  the  angle  which  a  straight  line  makes 
with  a  plane,'  it  is  always  the  angle  which  it  makes  with 
its  projection  on  the  plane  that  is  referred  to. 

448.  Properly  speaking,  two  straight  lines  form  an  angle 
only  when  they  intersect.  For  convenience,  however,  we  some- 
times speak  of  the  angle  between  two  non-intersecting  straight 
lines,  or  of  the  inclination  of  one  straight  line  to  another  which 
does  not  meet  it. 

By  the  angle  between  two  non-intersecting  straight  lines 
is  meant  the  angle  formed  by  two  intersecting  straight  lines 
which  are  parallel  to  them. 


444-449] 


LINES  AND  PLANES  IN  SPACE 


293 


Proposition  XX 

449.  The  acute  angle  which  a  straight  line  makes 
with  its  own  -projection  ujton  a  plane  is  the  least  angle 
it  makes  with  any  line  of  that  plane. 


Let  AB  be  any  straight  line  oblique  to  the  plane  L,  and  let 
BC  be  its  projection  upon  L. 

It  is  required  to  prove  that  /.ABC  is  less  than  the  angle 
which  AB  makes  with  any  other  line  of  the  plane. 

Proof.  From  any  point  A  of  the  line  AB  draw  the  perpen- 
dicular to  the  plane,  meeting  BC  at  C. 

From  B  draw  any  other  straight  line  BD  in  the  plane,  mak- 
ing the  segment  BD  equal  to  BC. 

Join  AD. 

Then  AD  is  greater  than  AC.     Why  ? 

Therefore  Z  ABD  is  greater  than  Z  ABC.  (Art.  80.) 

That  is,  Z  ABC  is  less  than  the  angle  which  AB  makes  with 
any  other  line  of  the  plane. 


EXERCISES 

1.  The  projection  of  any  straight  line  upon  a  given  plane  lies  in  a  plane 
through  the  given  line,  perpendicular  to  the  given  plane. 

2.  If  a  straight  line  is  parallel  to  a  given  plane,  the  projection  of  the 
line  upon  the  plane  is  parallel  to  the  line. 

3.  Parallel  lines  intersecting  the  same  plane  make  equal  angles  with  it. 


294 


ELEMENTARY  GEOMETRY 


[Chap.  VI 


Proposition  XXI 

450.  To  draw  a  straight  Uite  which  shall  he  perpen- 
dicular to  each  of  two  given  straight  lines  not  lying  in 
the  same  plane. 


Q 


R 


Let  a  and  b  be  two  straight  lines  not  lying  in  the  same  plane. 

It  is  required  to  find  a  straight  line  which  shall  be  perpen- 
dicular to  both  a  and  b. 

Construction.  Let  L  be  the  plane  through  a  which  is  par- 
allel to  b.  (Art.  414.) 

Also  let  c  be  the  projection  of  b  upon  the  plane  L. 

Then  c  is  parallel  to  b,  and  the  plane  of  b  and  c  is  perpen- 
dicular to  L.     Why  ? 

The  straight  lines  a  and  c  cannot  be  parallel,  for  then  a  and 
b  must  be  parallel,  contrary  to  hypothesis.  (Art.  396.) 

Let  a  and  c  intersect  at  P,  and  from  P  draw  PQ  perpen- 
dicular to  the  plane  L. 

Then  PQ  is  the  line  required. 

Proof.  Since  PQ  was  drawn  perpendicular  to  the  plane  L, 
at  the  point  P,  it  is  perpendicular  to  the  line  a,  and  lies  in  the 
plane  of  b  and  c.  (Art.  440.) 

And  since  PQ  is  perpendicular  to  the  line  c  [why?],  it  is 
also  perpendicular  to  b,  meeting  it,  say,  at  the  point  Q. 

Therefore  the  straight  line  PQ,  meeting  both  a  and  b,  is 
perpendicular  to  both. 


450-452]  LINES  AND  PLANES  IN   SPACE  295 

451.  Theorem.  Only  one  straight  line  can  be  drawn  perpen- 
dicular to  each  of  two  given  straight  lines  not  lying  in  the  same 
plane. 

If  possible  let  RS  be  a  second  straight  line  perpendicular  to 
both  a  and  h. 

Let  the  plane  of  h  and  RS  intersect  the  plane  L  in  the  line  d. 

Then  h  is  parallel  to  d  (Prop.  X),  and  RS  being  perpen- 
dicular to  h  is  also  perpendicular  to  d. 

Now  RS  was  assumed  perpendicular  to  a. 

Therefore  RS  is  perpendicular  to  the  plane  L.      (Prop.  IV.) 

But  RT  drawn  in  the  plane  of  h  and  c,  perpendicular  to  c,  is 
perpendicular  to  L.  (Prop.  XVIII.) 

Thus  we  have  two  perpendiculars  from  R  to  the  plane  L, 
which  is  impossible. 

Therefore  RS  cannot  be  perpendicular  to  both  a  and  6,  and 
similarly  no  other  line  than  PQ  can  be  perpendicular  to  both. 

452.  Theorem.  Tlie  common  perpendicular  to  two  given 
straight  lines  not  lying  in  the  same  plane,  is  the  shortest  line  be- 
tween them. 

The  shortest  line  from  a  point  of  a  to  the  line  b  is  perpen- 
dicular to  b,  and  the  shortest  line  from  a  point  of  b  to  the  line 
a  is  perpendicular  to  a. 

Therefore  the  shortest  line  from  a  point  of  a  to  a  point  of  b 
must  be  perpendicular  to  both  a  and  b. 

EXERCISES 

1.  Construct  a  plane  parallel  to  each  of  two  given  non-intersecting 
straight  lines  and  equidistant  from  them. 

2.  Draw  a  straight  line  to  intersect  three  given  non-intersecting  straight 
lines  at  P,  Q^  B,  respectively,  so  that  PQ  shall  equal  QP. 

3.  If  the  points  of  intersection  of  four  parallel  straight  lines  with  any- 
plane  form  the  vertices  of  a  parallelogram,  their  points  of  intersection 
with  every  plane  form  the  vertices  of  a  parallelogram. 


296  ELEMENTABY  GEOMETBY  [Chap.  VI 

Section  III 

POLYHEDRAL  ANGLES 

453.  Definition.  Three  planes  meeting  at  a  point  form 
what  is  called  a  trihedral  angle. 

The  planes  are  the  boundaries,  or  the  faces,  of  the  angle ;  the 
point  is  the  vertex  of  the  angle  ;  and  the  straight  lines  in  which 
the  planes  intersect,  two  and  two,  are  the  edges  of  the  angle. 

In  general,  the  figure  formed  by  several  planes  meeting  at 
a  point  is  called  a  polyhedral  angle.  The  planes  intersect, 
two  and  two,  in  a  certain  order  to  form  the  ^ 

edges. 

A  polyhedral  angle  may  be  designated  by 
naming  the  vertex  and  another  point  in  each 
edge  in  order.  Thus  the  polyhedral  angle  in 
the  diagram  may  be  designated  A-BCDE.       y 

A.  polyhedral  angle  of  four  faces  is  called  bL 
a  tetrahedral  angle.  '      ^l^ 

A  polyhedral  angle,  or  as  it  may  be  called,  a  solid  angle,  is  a  geometrical 
magnitude  of  a  character  essentially  different  from  a  plane  angle.  Like 
a  plane  angle,  however,  its  magnitude  depends,  not  upon  the  length  of  its 
edges,  but  only  upon  the  amount  of  their  divergence. 

454.  In  a  polyhedral  angle  the  pairs  of  adjacent  faces  form 
dihedral  angles,  and  the  two  edges  lying  in  any  face  form  a 
plane  angle. 

These  are  called  respectively  the  dihedral  angles  and  the 
face  angles  of  the  polyhedral  angle. 

The  dihedral  angles  and  the  face  angles  are  called  the  parts 
of  a  polyhedral  angle. 

A  polyhedral  angle  is  convex  when  the  section  of  its  faces 
made  by  any  plane  is  a  convex  polygon. 

455.  Two  polyhedral  angles  are  identically  equal  when  they 
can  be  made  to  coincide,  each  face  with  its  corresponding  face 
and  each  edge  with  its  corresponding  edge. 


453-457] 


LINES  AND  PLANES  IN  SPACE 


297 


456.  Two  polyhedral  angles  are  symmetrical  when  the  parts 
of  the  one  are  equal  respectively  to  the 

parts  of  the  other,  but  are  arranged  in 
reverse  order. 

Suppose  the  edges  of  any  polyhedral 
angle  0-ABCD  to  be  extended  through 
the  vertex,  to  points  A\  B',  C,  D'. 

The  polyhedral  angle  0-A'B'C'D' 
so  formed  is  symmetrical  to  the  angle 
0-ABCD.  For  the  face  angles  and 
the  dihedral  angles  are  equal,  each  to 
each,  being  vertically  opposite  in  each  case,  but  the  edges  are 
arranged  in  one  order  for  the  lower  angle,  and  in  the  opposite 
order  for  the  upper  angle. 

This  will  readily  be  seen  if  you  imagine  the  eye  placed  at  0, 
and  look  out  along  the  edges  in  order,  first  for  the  lower  angle, 
and  then  for  the  upper. 

457.  Two  symmetrical  polyhedral  angles  cannot  in  general 
be  made  to  coincide. 

Just  as  two  triangles  ABC  and  A'B'C  lying  in  the  same 
plane  may  be  identically 
equal,  but  cannot  be 
made  to  coincide  with- 
out turning  one  of  them 
over,  so  two  symmetri- 
cal polyhedral  angles 
can  in  general  be  made 
to  coincide  only  by 
turning  one  of  them 
inside  out,  so  to  speak,  by  drawing  the  vertex  down  through 
the  plane  section. 

In  the  same  way  one  of  your  hands  is  symmetrical  to  the 
other,  not  identically  equal  to  it.  The  one  hand  cannot  be  made 
to  occupy  exactly  the  same  space  as  the  other,  and  the  left  glove 
will  fit  the  right  hand  only  when  it  has  been  turned  inside  out. 


298  ELEMENTARY  GEOMETRY  [Chap.  VI 


Proposition  XXII 

458.   The  sum  of  any  two  face  angles  of  a  trihedral 
angle  is  greater  than  the  third  face  angle. 


Let  A-BCD  be  any  trihedral  angle  in  which,  the  face  angle 
BAD  is  greater  than  either  of  the  other  two. 

It  is  required  to  prove  that  the  sum  of  the  face  angles  BAC 
and  CAD  is  greater  than  the  face  angle  BAD. 

Proof.  In  the  face  BAD  draw  the  straight  line  AE  making 
Z  BAE  equal  to  Z  BAC. 

Make  the  line-segments  AE  and  AC  equal,  and  through  C 
and  E  pass  a  plane  cutting  the  other  two  edges  at  B  and  D. 

Then  BE  =  BC.  (Art.  41.) 

Therefore  CD  is  greater  than  ED,  since  BC  and  CD  are 
together  greater  than  BD.  (Art.  70.) 

In  As  CAD  and  EAD,  AC=AE,  AD  =  AD,  while  CD  is 
greater  than  ED. 

Therefore       Z  CAD  is  greater  than  Z  EAD.  (Art.  80.) 

Therefore  the  sum  of  Zs  BAC  and  CAD  is  greater  than 
ZBAD. 

459.  Corollary.  Any  face  angle  of  a  polyhedral  angle  is 
less  than  the  sum  of  the  remaining  face  angles. 


458-460] 


LINES  AND  PLANES  IN  SPACE 


299 


Proposition  XXIII 

460.  The  sum  of  the  face  angles  of  any  convex  poly- 
hedral angle  is  less  than  four  right  angles. 


Let  A-BCDE  be  any  convex  polyhedral  angle. 

It  is  required  to  prove  that  the  sum  of  the  face  angl.es  BAC, 
CAD,  DAE,  etc.,  is  less  than  four  right  angles. 

Proof.  Choose  any  plane  cutting  all  the  faces  and  edges  of 
the  given  polyhedral  angle.  The  section  will  be  a  convex 
polygon  BCDE.     Why  ? 

The  interior  angles  of  the  polygon  BCDE,  together  with 
four  right  angles,  make  twice  as  many  right  angles  as  the  poly- 
gon has  sides.  (Art.  116.) 

In  any  triangular  face,  as  A  BAC,  the  sum  of  the  angles  is 
two  right  angles.  (Art.  101.) 

Hence,  the  sum  of  all  the  angles  in  all  the  triangular 
faces  equals  twice  as  many  right  angles  as  the  polyhedral 
angle  has  faces,  i.e.,  twice  as  many  right  angles  as  the  polygon 
BCDE  has  sides. 

The  angles  in  all  the  triangular  faces  may  be  grouped  as  the 
face  angles  at  the  vertices,  B,  C,  D,  etc.,  together  with  the  face 
angles  at  A. 

Therefore,  the  face  angles  at  B,  C,  D,  etc.,  -f-  the  face  angles 
at  J.  =  the  angles  of  the  polygon  BCDE  +  four  right  angles, 


300 


ELEMENTARY  GEOMETRY 


[Chap.  VI 


since  each  of  these  sums  equals  twice  as  many  right  angles  as 
the  polygon  BCDE  has  sides. 

Now  at  any  vertex,  as  B,  the  two  face  angles,  ABC  and 
ABE,  are  together  greater  than  the  angle  of  the  polygon  EBG. 

(Prop.  XXII.) 

Hence  the  sum  of  the  face  angles  at  all  the  vertices,  B,  C,  D, 
etc.,  is  greater  than  the  sum  of  the  angles  of  the  polygon  BCDE. 

Therefore,  the  sum  of  the  face  angles  at  A  must  be  less 
than  four  right  angles. 


Proposition  XXIV 

461.  If  two  trihedral  angles  have  the  three  face  angles 
of  one  equal,  respectively,  to  the  three  face  angles  of  the 
other,  their  corresponding  dihedral  angles  are  also  equal. 


Let  A-BCD  and  A'-B'C'D'  be  two  trihedral  angles  having 
the  face  angles  BAC,  CAD,  DAB  equal,  respectively,  to  the 
face  angles  BA'C,  C'A'D',  D'A'B'. 

It  is  required  to  prove  that  the  dihedral  angles  whose  edt^es 
are  AB,  AC,  AD  are  equal,  respectively,  to  the  dihedral  angles 
whose  edges  are  A'B',  A'C,  A'D'. 

Proof.  Let  the  planes  BCD  and  B'C'D'  cut  the  edges  of  the 
two  trihedral  angles  so  as  to  make  the  segments  AB,  AC,  AD, 
A'B',  A'C,  A'D'  all  equal. 

In  AB  and  A'B'  choose  two  points,  P  and  P',  equidistant 
from  A  and  A',  and  through  these  points  pass  planes  perpen- 
dicular to  the  edges  AB  and  A'B'. 


460-461]  LINES  AND  PLANES  IN   SPACE  301 

The  plane  through  P  perpendicular  to  AB  will  intersect  BC 
and  BD  at  some  points  Q  and  K.     Why  ? 

Notice  that  Q  and  B  are  not  necessarily  within  the  segments 
BC  and  BD. 

Join  PQ,  PjR,  and  QR. 

Similarly  construct  the  triangle  P'Q'E'. 

Zs  QPR  and  Q'P'Ii'  are  the  plane  angles  of  the  dihedral 
angles  whose  edges  are  AB  and  A'B'  [why?],  and  it  is 
required  to  prove  them  equal. 

As  ABC  and  A'B'C  are  identically  equal  isosceles  triangles, 
by  hypothesis  and  construction. 

As  QBP  and  Q'B'P'  are  identically  equal.  (Art.  43.) 

Therefore  BQ  =  B'Q', 

and  PQ  =  PQ'. 

Similarly  from  As  EBP  and  P'B'P, 
BR  =  B'R'. 
PR  =  PR'. 
Now  As  BCD  and  B'OD'  are  identically  equal.        (Art.  53.) 

Point  out  all  the  steps  necessary  to  show  this. 
Therefore  .      Z  CBD  =  Z  OB'D\ 

Therefore  As  QBR  and  Q'B'R'  are  identically  equal  (Art. 41), 
and  QR  =  Q'R'. 

Therefore  As  QPR  and  Q'P'R'  are  identically  equal  (Art.  53), 
and  Z  QPR  =  Z  Q'PR'. 

That  is,  the  dihedral  angle  whose  edge  is  AB  equals  the 
dihedral  angle  whose  edge  is  A'B'. 

Similarly  it  may  be  shown  that  the  other  dihedral  angles 
are  equal. 


302 


ELEMENTARY  GEOMETRY 


[Chap.  VI 


Note.  If  the  parts  of  the  trihedral  angles  A-BCD  and  A'-B'C'D' 
are  arranged  in  the  same  order  around  A^  the  trihedral  angles  are  equal, 
and  if  in  the  reverse  order,  they  are  symmetrical  but  not  equal.  In  the 
first  case,  they  could  be  superposed  ;  in  the  second  case,  they  could  not 
be  superposed. 

The  pupil  should  try  to  superpose  the  one  trihedral  angle  on  the  other, 
beginning  with  the  vertices  A  and  A'  and  the  equal  face  angles  BAC  and 
B'A'C.  Will  the  edges  AD  and  A'D'  then  lie  on  the  same  side  of  this 
face? 

If  the  edge  AB  is  made  to  coincide  with  its  corresponding  edge  A'B'^ 
and  AC  with  A'C,  how  will  AD  and  A'D<  lie  ? 

If  AB  is  made  to  coincide  with  A'C  and  AC  with  A'B',  how  will  AD 
and  A'D'  lie  ? 

462.  Definition.    An   isosceles  trihedral  angle   is   one    in 

which  two  of  the  face  angles  are  equal. 

463.  Theorem.  An  isosceles  trihedral  angle  and  its  symmet- 
rical trihedral  angle  are  identically  equal,  i.e.  can  be  superposed. 

Suppose  that  A-BCD  is  an  isosceles  trihedral  angle  havin^;^ 
the  face  angles  BACsmd  BAD  equal ;  and  suppose  that  A'-B'C'D' 
is  a  trihedral  angle  having  its 
face  angles  respectively  equal 
to  those  of  A-BCD,  but  ar- 
ranged in  reverse  order. 

The  corresponding  dihedral 
angles  are  equal  by  Proposition 
XXiy.  Therefore  the  dihedral 
angle  whose  edge  is  AB  is  equal 
to  the  dihedral  angle  whose  edge  is  A'B'. 

Place  the  trihedral  angle  A'-B'C'D'  upon  the  trihedral 
angle  A-BCD  so  that  the  dihedral  angle  A'B'  coincides  with 
the  dihedral  angle  AB,  the  vertex  A'  coinciding  with  the 
vertex  A. 

Then  since  Z  B'A'D'  =  Z  BAD  =  Z  BAC,  by  hypothesis,  the 
edge  A'D'  will  coincide  with  the  edge  AC 

Similarly,  the  edge  A'C  will  coincide  with  the  edge  AD, 
and  the  two  trihedral  angles  will  coincide  in  all  their  parts. 


461-467]  LINES  AND  PLANES  IN  SPACE  303 

Why  could  not  two  symmetrical  trihedral  angles  whose  face 
angles  are  all  unequal  be  superposed  in  the  same  way?  At 
what  point  would  the  process  of  superposition  fail  ? 

464.  Corollary  I.  If  two  face  angles  of  a  trihedral  angle 
are  equals  the  dihedral  angles  opposite  them  are  also  equal. 

For,  the  trihedral  angle  and  its  symmetrical  trihedral  angle 
can  be  superposed. 

465.  Corollary  II.  If  two  isosceles  trihedral  angles  have 
the  three  face  ayigles  of  one  equal,  respectively,  to  the  three  face 
angles  of  the  other,  they  are  identically  equal. 

They  can  be  superposed  whether  their  parts  are  arranged  in 
the  same  order  or  in  opposite  orders. 

466.  Theorem.  If  two  trihedral  ayigles  have  two  face  angles 
and  the  included  dihedral  arigle  of  one  equal,  respectively,  to  two 
face  angles  and  the  included  dihedral  angle  of  the  other,  they 
are  either  identically  equal  or  symmetrical,  according  as  the  parts 
are  arranged  in  the  same  order  or  in  opposite  orders. 

n  the  parts  are  arranged  in  the  same  order,  the  two  can  be 
superposed  ;  if  arranged  in  opposite  orders,  one  can  be  super- 
posed to  the  symmetric  of  the  other, 

467.  Theorem.  If  two  trihedral  angles  iiave  a  face  angle 
and  the  two  adjacent  dihedral  angles  of  one  equal,  respectively, 
to  a  face  angle  and  the  two  adjacent  dihedral  angles  of  the  other, 
they  are  either  identically  equal  or  symmetrical,  according  as  the 
parts  are  arranged  in  the  same  order  or  in  opposite  orders. 

EXERCISES 

1.  The  planes  that  bisect  the  dihedral  angles  of  any  trihedral  angle 
intersect  in  the  same  straight  line. 

2.  If  the  face  angle  AOB  of  the  trihedral  angle  0-ABC  is  bisected  by 
the  straight  line  OD^  the  angle  COD  is  less  than  half  the  sum  of  the  face 
angles  ^OC  and  ^00. 


304  ELEMENTARY  GEOMETRY  [Chap.  VI 


MISCELLANEOUS  EXERCISES 

1.  From  a  point  A  in  one  of  two  intersecting  planes,  AB  is  drawn 
perpendicular  to  the  first  plane,  meeting  the  second  plane  at  B,  and  AC 
is  drawn  perpendicular  to  the  second  plane,  meeting  it  at  C.  Show  that 
^O  is  perpendicular  to  the  line  of  intersection  of  the  two  planes. 

2.  Two  line  segments  AB  and  DC  are  such  that  if  BC  and  AD  are 
joined  each  of  the  angles  A,  B,  C,  D,  is  a  right  angle.  Prove  that  AB 
and  DC  are  parallel. 

3.  If  two  triangles  ABC  and  A'B'C  in  different  planes  are  so  situ- 
ated that  the  lines  AA',  BB\  CC  meet  in  a  point  S^  then  the  pairs  of 
corresponding  sides  AB,  A'B' ;  AC,  A'C  ;  BC,  B' C  intersect  in  points 
of  one  straight  line. 

Suggestion.  Let  the  planes  of  ABC  and  A'B'C  intersect  in  a  straight 
line  s.  The  points  S,  A,  A',  B,  B'  lie  in  one  plane  intersecting  s  at  X, 
say.     AB  and  A'B'  must  intersect  at  X.     Similarly  for  the  other  pairs. 

4.  Show  that  the  locus  of  a  point  equidistant  from  three  given  points 
is  a  straight  line  through  the  centre  of  the  circle  determined  by  the  three 
points. 

5.  Prove  that  if  AP  makes  equal  angles  with  AB  and  AC,  it  must  lie 
in  one  or  other  of  two  fixed  planes. 

6.  Of  three  given  planes  each  is  perpendicular  to  the  other  two.  Show 
that  of  their  three  lines  of  intersection  each  is  perpendicular  to  the  other 
two. 

7.  If  two  intersecting  planes  are  cut  by  two  parallel  planes,  the  lines 
of  section  of  the  first  two  by  each  of  the  parallel  planes  will  make  equal 
angles. 

8.  If  a  straight  line  is  equally  inclined  to  two  given  planes,  show  that 
it  is  perpendicular  to  the  plane  bisecting  one  of  the  dihedral  angles  formed 
by  the  two  planes. 

9.  Show  that  the  locus  of  the  foot  of  the  perpendiculars  drawn  from 
a  fixed  point  to  the  planes  through  a  fixed  straight  line  is  a  circle. 

10.  Draw  through  the  vertex  of  a  trihedral  angle  a  straight  line  which 
will  make  equal  angles  with  the  three  edges  of  the  trihedral  angle.  How 
many  such  lines  are  there  ? 

11.  The  projections  qp  a  plane  of  two  equal  and  parallel  line-segments 
are  equal  and  parallel. 


Summary]  LINES  AND   PLANES  IN   SPACE  305 

SUMMARY  OF   CHAPTER  VI 
1.    Definitions. 

(1)  Solid  Figure  —  one  whose  points  and  lines  do  not  all  lie  in  the 

same  plane.     §  386. 

(2)  Line  Parallel  to  a  Plane  —  a  straight  line  which  does  not  meet 

the  plane,  however  far  they  may  be  extended.     §  392. 

(3)  Line  Perpendicular  to  a  Plane  — a,  straight  line  which  is  perpen- 

dicular to  every  line  of  the  plane  passing  through  their  inter- 
section.    §  393. 

(4)  Line  Oblique  to  a  Plane — a  straight  line  neither  parallel  nor 
^  perpendicular  to  the  plane.     §  393. 

(5)  Parallel  Planes  —  planes  which  do  not  meet,  however  far  they 

may  be  extended.     §  393. 

(6)  Skeiv  or  Gauche  Lines  —  straight  lines  so  situated  that  no  plane 

can  contain  them  both.     Ex.  3,  p.  271. 

(7)  Distance  from  a  Point  to  a  Plane  —  the  length  of  the  perpen- 

dicular from  the  point  to  the  plane.     §  411. 

(8)  Dihedral  Angle  —  the  figure  formed  by  two  planes  meeting  in  a 

straight  line.     §  426. 

(9)  Bight  Dihedral  Angle  — an  angle  formed  by  two  intersecting 

planes  when  the  adjacent  angles  so  formed  are  equal.     §  429. 

(10)  Plane  Angle  of  a  Dihedral  Angle  —  the  angle  between  two  straight 

lines  drawn  perpendicular  to  the  edge  from  the  same  point,  one 
in  each  boundary,     §  430. 

(11)  Projection  of  a  Point  upon  a  Plane  —  the  foot  of  the  perpen- 

dicular drawn  from  the  point  to  the  plane.     §  445. 

(12)  Projection  of  a  Line  upon  a  Plane  —  the  locus  of  the  projections 

of  its  points  upon  the  plane.     §  445. 

(13)  Tnclination  of  a  Line  to  a  Plane  —  the  angle  between  the  line 

and  its  projection  upon  the  plane.     §  447. 

(14)  Trihedral  Angle  — the  figure  formed  by  three  planes  meeting  at 

a  point.     §  453. 

(15)  Polyhedral  Angle  —  the  figure  formed  by  several  planes  meeting 

at  a  point.     §  453. 

(16)  Identically  Equal  Polyhedral  Angles  —  two  which  can  be  made 

to  coincide.     §  455. 

(17)  Symmetrical  Polyhedral  Angles  —  two  whose  parts  are  equal, 

respectively,  but  arranged  in  reverse  order.     §  456. 

(18)  Isosceles  Trihedral  Angle  —  one  having  two  face  angles  equal. 

§462. 

X 


306  ELEMENTARY   GEOMETRY  [Chap.  VI 

2.    Postulate. 

(1)  Through  three  points  not  in  the  same  straight  line  there  can 
pass  one  and  only  one  plane  (Postulate  9).     §  388. 


^/ 


^.    Elements  which  determine  a  Plane. 

(1)  Three  points  (Postulate  9).     §  388. 

(2)  A  straight  line  and  a  point  not  lying  on  it.     §  388. 

(3)  Two  intersecting  straight  lines.     §  388. 

(4)  Two  parallel  straight  lines.     §  388. 

4.  Problems. 

(1)  At  a  given  point  in  a  given  plane  to  erect  a  perpendicular  to  the 

plane.     §  403. 

(2)  From  a  given  point  without  a  given  plane  to  draw  a  perpendicular 

to  the  plane.     §  408. 

(3)  To  draw  a  straight  line  which  shall  be  perpendicular  to  each  of 

two  given  straight  lines  not  lying  in  the  same  plane.     §  450. 

5.  Theorems  on  the  Intersections  of  Planes. 

(1)  The  intersection  of  two  planes  is  a  straight  line.     §  390. 

(2)  If  two  planes  have  one  point  in  common,  they  must  have  a  second 

point,  and  hence  a  straight  line,  in  common.     §  391. 

(3)  If  three  planes  intersect,  two  and  two,  their  three  lines  of  inter- 

section are  either  concurrent,  or  are  parallel,  two  and  two. 
§  394. 

(4)  The  two  planes  determined  by  two  given  parallel  lines  and  a 

point  not  lying  in  their  plane,  intersect  in  a  line  parallel  to 
each  of  the  given  lines.     §  395. 

(5)  Two  parallel  planes  are  intersected  by  any  third  plane  in  parallel 

lines.     §  419. 

6.  Theorems  on  Straight  Lines  perpendicular  to  Planes. 

(1)  If  a  straight  line  is  perpendicular  to  each  of  two  given  straight 

lines  at  their  point  of  intersection,  it  is  perpendicular  to  the 
plane  of  these  lines.     §  397. 

(2)  At  any  point  of  a  straight  line  one  plane  can  be  constructed  per- 

pendicular to  that  line,  and  only  one.     §  398. 

(3)  Through  a  given  point  not  on  a  given  straight  line,  one  plane  and 

only  one  can  be  constructed  perpendicular  to  the  given  line. 
§  399. 

(4)  Two  intersecting  planes  cannot  both  be  perpendicular  to  the  same 

straight  line.     §  400. 


Summary]  LINES  AND  PLANES  IN  .SPACE  307 

(5)  All  the  straight  lines  perpendicular  to  a  given  line  at  a  given 

point  lie  in  one  plane  perpendicular  to  the  given  line.     §  401. 

(6)  If  a  plane  bisects  a  given  line-segment  perpendicularly,  every 

point  of  the  plane  is  equidistant  from  the  extremities  of  the 
line-segment,  and  conversely.     §  402. 

(7)  At  a  point  in  a  plane  but  one  straight  line  can  be  drav^n  perpen- 

dicular to  the  plane.     §  404. 

(8)  From  a  point  without  a  plane  only  one  perpendicular  to  the  plane 

can  be  drawn.     §  409. 

(9)  Two  straight  lines  perpendicular  to  the  same  plane  are  parallel. 

§405. 

(10)  If  one  of  two  parallel  lines  is  perpendicular  to  a  plane,  the  other 

is  also.     §  407. 

(11)  Two  planes  perpendicular  to  the  same  straight  line  are  parallel. 

§412. 

(12)  A  straight  line  perpendicular  to  one  of  two  parallel  planes  is  also 

perpendicular  to  the  other.     §  422. 

(13)  If  a  straight  line  is  perpendicular  to  a  given  plane,  every  plane 

containing  that  line  is  perpendicular  to  the  given  plane.    §  436. 

7.  Theorems  on  Straight  Lines  and  Planes  parallel  to  them. 

(1)  If  two  straight  lines  are  parallel,  any  plane  containing  one  of 

them,  and  not  the  other,  is  parallel  to  the  other.     §  413. 

(2)  Through  either  of  two  given  straight  lines  not  lying  in  the  same 

plane,  one  plane  can  be  passed  parallel  to  the  other  line.    §  414. 

(3)  Through  "a  given  point  a  plane  can  be  passed  parallel  to  any  two 

given  straight  lines  in  space.     §  415. 
- — ^(4)  If  a  straight  line  is  parallel  to  a  given  plane,  it  is  parallel  to  the 
intersection  of  any  plane  through   it,  with  the  given  plane. 
§  416. 

(5)  If  a  straight  line  is  parallel  to  a  given  plane,  a  line  drawn  from 

any  point  in  the  plane  parallel  to  the  given  line  lies  in  the  given 
plane.    §  417. 

(6)  If  two  intersecting  straight  lines  are  each  parallel  to  a  given  plane, 

the  plane  determined  by  these  lines  is  also  parallel  to  the  given 
plane.     §  418. 

8.  Theorems  on  Planes  perpendicular  to  Each  Other. 

(1)  If  a  straight  line  is  perpendicular  to  a  given  plane,  every  plane 

containing  that  line  is  perpendicular  to  the  given  plane.    §  436. 
(2)  Any  plane  perpendicular  to  the  edge  of  a  dihedral  angle  is  per- 
pendicular to  each  of  its  faces.     §  438. 


308  ELKMENTARY  GEOMETRY  [Chap.  VI 

(3)  If  two  planes  are  perpendicular  to  each  other,  a  straight  line 

drawn  in  one  of  them,  perpendicular  to  their  intersection,  is 
perpendicular  to  the  other.     §  439. 

(4)  If  two  planes  are  perpendicular  to  each  other,  a  straight  line 

drawn  from  any  point  of  their  intersection,  perpendicular  to 
one  plane,  must  lie  in  the  other.     §  440. 

(5)  If  two  planes  are  perpendicular  to  each  other,  a  straight  line 

drawn  from  any  point  in  one  of  them,  perpendicular  to  the 
other,  must  lie  in  the  first  plane.     §  442. 

(6)  If  two  intersecting  planes  are  each  perpendicular  to  a  third  plane, 

their  intersection  is  also  perpendicular  to  that  plane.    §  443. 

9.  Theorems  on  Parallel  Planes. 

^ (1)  Two  parallel  planes  are  intersected  by  any  third  plane  in  parallel 

lines.     §  419. 

(2)  Parallel  line-segments  terminated  by  parallel  planes  are  equal. 

§420. 

(3)  Two  parallel  planes  are  everywhere  equidistant.     §  421. 

(4)  A  straight  line  perpendicular  to  one  of  two  parallel  planes  is  also 

perpendicular  to  the  other.     §  422. 

(5)  If  two  straight  lines  are  cut  by  three  parallel  planes,  the  corre- 

sponding segments  are  proportional.     §  425. 

10.  Theorems  on  Dihedral  Angles. 

(1)  All  plane  angles  of  the  same  dihedral  angle  are  equal.     §  431. 
•  (2)  The  lines  of  intersection  with  the  boundaries  of  a  dihedral  angle, 
of  any  plane  perpendicular  to  the  edge  of  that  angle,  form  the 
plane  angle  of  the  dihedral  angle.     §  432. 

(3)  Two  dihedral  angles  are  equal  if  their  plane  angles  are  equal. 

§  433. 

(4)  Two  dihedral  angles  are  in  the  same  ratio  as  their  plane  angles. 

§  434. 

(5)  The  locus  of  points  equidistant  from  the  boundaries  of  a  dihedral 

angle  is  the  plane  bisecting  that  angle.     §  444. 

11.  Theorems  on  Trihedral  and  Polyhedral  Angles. 

(1)  The  sum  of  any  two  face  angles  of  a  trihedral  angle  is  greater 

than  the  third  face  angle.     §  458. 

(2)  Any  face  angle  of  a  polyhedral  angle  is  less  than  the  sum  of  the 

remaining  face  angles.     §  459. 

(3)  The  sum  of  the  face  angles  of  any  convex  polyhedral  angle  is  less 

than  four  right  angles.     §  460. 


Summary]  LINES  AND  PLANES  IN   SPACE  309 

(4)  If  two  trihedral  angles  have  the  three  face  angles  of  one  equal, 

respectively,  to  the  three  face  angles  of  the  other,  their  corre- 
sponding dihedral  angles  are  also  equal.     §  461. 

(5)  An  isosceles  trihedral  angle  and  its  symmetrical  trihedral  angle 

are  identically  equal.     §  4G;3. 

(6)  If  two  face  angles  of  a  trihedral  angle  are  equal,  the  dihedral 

angles  opposite  them  are  also  equal.     §  464. 

(7)  If  two  isosceles  trihedral  angles  have  the  three  face  angles  of  one 

equal,  respectively,  to  the  three  face  angles  of  the  other,  they 
are  identically  equal.     §  465. 

(8)  If  two  trihedral  angles  have  two  face  angles  and  the  included 

dihedral  angle  of  one  equal,  respectively,  to  two  face  angles  and 
the  included  dihedral  angle  of  the  other,  they  are  either  iden- 
tically equal  or  symmetrical,  according  as  the  parts  are  arranged 
in  the  same  order  or  in  opposite  orders.     §  466. 

(9)  If  two  trihedral  angles  have  a  face  angle  and  the  two  adjacent 

dihedral  angles  of  one  equal,  respectively,  to  a  face  angle  and 
the  two  adjacent  dihedral  angles  of  the  other,  they  are  either 
identically  equal  or  symmetrical,  according  as  the  parts  are 
arranged  in  the  same  order  or  in  opposite  orders.     §  467. 

12,    Miscellaneous  Theorems. 

(1)  Two  straight  lines  each  parallel  to  a  third  line  are  parallel  to  each 

other.     §  396. 

(2)  If  from  the  foot  of  a  given  perpendicular  to  a  plane,  a  straight 

line  is  drawn  at  right  angles  to  any  line  of  the  plane,  any  line 
through  their  intersection  which  meets  the  given  perpendicular 
is  at  right  angles  to  the  line  of  the  plane.     §  406. 

(3)  Of  all  straight  lines  which  can  be  drawn  from  a  point  to  a  plane, 

the  perpendicular  is  the  shortest.     §  410. 

(4)  If  two  intersecting  straight  lines  lying  in  one  plane  are  parallel, 

respectively,  to  two  intersecting  straight  lines  lying  in  another 
plane,  the  two  planes  must  be  parallel,  and  the  angles  formed 
by  the  lines  are  equal.     §  424. 

(5)  The  acute  angle  which  a  straight  line  makes  with  its  own  projec- 

tion upon  a  plane  is  the  least  angle  it  makes  with  any  line  of 
that  plane.     §  449. 

(6)  Only  one  straight  line  can  be  drawn  perpendicular  to  each  of  two 

given  straight  lines  not  lying  in  the  same  plane.     §  451. 

(7)  The  common  perpendicular  to  two  given  straight  lines  not  lying 

in  the  same  plane  is  the  shortest  line  between  them.     §  452. 


CHAPTER  VII 
PRISMS  AND  PYRAMIDS 

Section  I 

AREA  AND  VOLUME  OF  A  PRISM 
Definitions 

468.  A  section  of  a  surface  made  by  an  intersecting  plane 
is  the  locus  of  points  common  to  the  surface  and  the  plane. 

469.  A  surface,  such  that  every  section  of  it  made  by  an 
intersecting  plane  consists  of  one  or  more  closed  lines,  is  called 
a  closed  surface. 

470.  A  closed  surface  which  is  made  up  wholly  of  intersect- 
ing planes  is  called  a  polyhedron. 

The  planes  are  called  the  faces  of 
the  polyhedron ;  the  lines  in  which 
the  faces  intersect  are  called  the 
edges;  the  points  in  which  the  edges 
intersect  are  called  the  vertices. 

The  line-segment  connecting  any  two  vertices  not  lying  in 
the  same  face  is  called  a  diagonal  of  the  polyhedron. 

If  a  polyhedron  lies  wholly  on  one  side  or  the  other  of  each 
of  its  faces,  it  is  called  a  convex  polyhedron. 

Any  section  of  a  convex  polyhedron  made  by  a  plane  is  a 
convex  polygon. 

310 


4-f 


468-475] 


PRISMS  AND  PYRAMIDS 


311 


A  polyhedron  of  four  faces  is  called  a  tetrahedron  ;  one  of 
five  faces  is  called  a  pentahedron ;  one  of  six  faces  is  called  a 
hexahedron ;  one  of  eight  faces  is  called  an  octahedron ;  one  of 
twelve  faces  is  called  a  dodecahedron ;  one  of  twenty  faces  is 
called  an  icosahedron. 

471.   A  polyhedron  of  which  two  faces  are  convex  polygons 
lying  in  parallel  planes  and  identically 
equal,  while  the  remaining  faces  are 
parallelograms,  is  called  a  prism. 

The  equal  parallel  faces  are  called 
the  bases  of  the  prism;  the  remaining 
faces,  the  lateral  faces.  The  edges 
lying  in  the  bases  are  called  the  base 

edges ;  and  the  intersections  of   the  lateral  faces,  the  lateral 
edges. 

The  base  edges  of  a  prism  are  equal  and  parallel,  two  and 
two,  each  edge  of  one  base  being  equal  and  parallel  to  an  edge 
of  the  other  base. 

The  lateral  edges  of  a  prism  are  all  equal  and  parallel 
(Art.  420),  and  make  equal  angles  with  the  plane  of  either  base. 

A  prism  is  called  triangular,  quadrilateral,  etc.,  according  as 
its  bases  are  triangles,  quadrilaterals,  etc. 


472.  A  right  prism  is  one  whose  lateral  edges  are  perpen- 
dicular to  its  bases. 

If  the  lateral  edges  are  not  perpendicular  to  the  bases,  the 
prism  is  called  oblique. 

473.  A  right  section  of  a  prism  is  the  section  made  by  any 
plane  perpendicular  to  the  lateral  edges. 

474.  The  lateral  area  of  a  prism  is  the  sum  of  the  areas  of 
the  lateral  faces  of  the  prism. 


475.   The  altitude  of  a  prism  is  the  perpendicular  distance 
between  its  bases. 


312 


ELEMENTARY  GEOMETRY  [Chap.  VII 


Proposition  I 

476.   The  sections  of  a  prism  made  hy  parallel  planes 
are  polygons  which  are  identically  equal. 


Let  the  sections  of   the   prism  PQ,  made  by  the  parallel 
planes  L  and  M,  be  ABODE  and  A'B'C'D'E'. 

It   is   required    to   prove   that   the   polygons   ABODE  and 
A'B'O'D'E'  are  identically  equal. 

Proof.     First,        AB  is  parallel  to   A'B'.  (Art.  419.) 

And  since  AA'  is  parallel  to  BB'  by  definition,  AA'B'B  is  a 
parallelogram.     Therefore  AB  =  A'B'. 

Similarly,  BO=B'0',  OD  =  O'D,  and  so  on. 

Hence,    ABODE  and  A'B'O'D'E'  are  mutually  equilateral. 

Next,  since  AB  is  parallel  to  A'B',  and  BO  is  parallel  to  B'O', 

Z  ABO  =  Z  A'B'O.  (Art.  424.) 

Similarly,  Z  BOD=ZB'0'D',  Z  ODE=Z  O'D'E',  and  soon. 
Hence,    ABODE  and  A'B'O'D'E'  are  mutually  equiangular. 
Therefore  the  polygons  ABODE  and  A'B'O'D'E'  are  identi- 
cally equal. 

Note.     The  same  proof  applies  if  the  planes  of  section  intersect  some 
or  all  of  the  edges  produced. 


476-480]  PRISMS  AND  PYRAMIDS  313 

477.  Corollary  I.  All  right  sections  of  the  same  prism  are 
identically  equal. 

478.  Corollary  II.  The  section  of  a  prism  made  by  a  plane 
parallel  to  either  base  is  identically  equal  to  the  base. 

Proposition  II 

479.  The  lateral  area  of  a  prism  is  equal  to  the  prod- 
uct of  a  lateral  edge  and  the  perimeter  of  a  right  sec- 
tion of  the  prism. 

Q 


Let  ABCDE  be  a  right  section  of  the  prism  PQ,  and  PR  be 
any  one  of  the  lateral  edges. 

It  is  required  to  prove  that  the  lateral  area  of  PQ  is  equal  to 
the  product  of  PR  and  the  perimeter  of  ABCDE. 

Proof.     AB  is  perpendicular  to  PR.     Why  ? 

Therefore  the  area  of  the  lateral  face  PS  is  equal  to  the 
product  of  PR  and  AB.  (Art.  305.) 

Similarly,  the  area  of  each  lateral  face  is  equal  to  the 
product  of  a  lateral  edge  (=  PR)  and  a  side  of  ABCDE. 

Therefore  the  lateral  area  of  PQ  is  equal  to  the  product  of 
PR  and  {AB  ^  BC  +  CD  +  etc.). 

That  is,  the  lateral  area  of  PQ  is  equal  to  the  product  of 
PR  and  the  perimeter  of  ABCDE. 

480.  Corollary.  The  lateral  area  of  a  right  prism  is  equal 
to  the  product  of  its  altitude  and  the  perimeter  of  its  base. 


314  ELEMENTARY  GEOMETRY  [Chap.  VII 


Proposition  III 

481.  Two  prisms  are  identically  equal  if  the  three 
faces  forming  a  trihedral  angle  in  one  are  identically 
equal  to  the  three  faces  forming  a  trihedral  angle  in 
the  other  and  are  similarly  placed. 


Let  AJ  and  A'J^  be  two  prisms  having  the  three  faces  BE, 
BF,  and  BII,  forming  the  trihedral  angle  B,  identically  equal, 
respectively,  to  the  three  faces  B'E',  B'F',  B'H',  forming  the 
trihedral  angle  B',  the  equal  faces  being  similarly  placed. 

It  is  required  to  prove  that  the  two  prisms  AJ  and  A' J'  are 

identically  equal. 

Proof.  First,  the  trihedral  angles  B  and  B'  are  identically 
equal.  (Art.  461.) 

Apply  the  prism  A' J'  to  the  prism  A  J  so  that  the  trihedral 
angle  B'  coincides  with  the  trihedral  angle  B. 

Then,  since  the  faces  forming  these  angles  are  identically 
equal,  two  and  two,  the  vertices  A',  B',  C,  D',  E',  F',  G',  H', 
will  coincide  with  the  vertices  A,  B,  C,  D,  E,  F,  G,  II',  and  the 
edges  connecting  these  vertices  will  also  coincide. 

Now,  since  the  lateral  edges  of  each  prism  are  parallel  and 
equal  (Art.  471),  and  the  edges  A'F',  B'G',  C'H',  coincide 
respectively  with  the  edges  AF,  BG,  CH,  while  the  vertices 
/>',  E',  coincide  with  D,  E,  therefore  the  remaining  lateral 
edges  D'J'  and  E'K'  must  coincide  with  the  remaining  edges 


481-487]  PRISMS  AND  PYRAMIDS  315 

DJ  and  EK^  and  the  remaining  vertices  J'  and  K'  with  the 
remaining  vertices  J  and  K. 

Hence  the  two  prisms  coincide  throughout. 

482.   Corollary  I.     Two  right  prisms  are  identically  equal 
if  their  bases  are  identically  equal  and  they  have  equal  altitudes. 

If  the  sides  of  the  bases  are  not  arranged  in  the  same  order, 
one  of  the  prisms  may  be  turned  over,  and  then  they  can  be 
made  to  coincide. 

Could  this  be  done  if  the  prisms  were  not  right  prisms  ? 

Why  is  it  necessary  to  have  the  words  'and  similarly  placed' 
in  the  enunciation  of  the  main  theorem  ? 


483.  Definition.  A  truncated  prism  is  a  poly- 
hedron having  parallel  lateral  edges  like  a  prism, 
but  bases  which  are  neither  parallel  nor  equal. 

The  lateral  faces  cannot  all  be  parallelograms, 
though  some  of  them  may  be. 


^^ 


484.  Theorem.  Two  truncated  prisms  are  identically  equal 
if  the  three  faces  forming  a  trihedral  angle  in  one  are  identically 
equal  to  the  three  faces  forming  a  trihedral  angle  in  the  other ,  and 
are  similarly  placed. 

485.  Definition.  The  space  contained  within  a  closed  sur- 
face is  called  the  volume  of  the  figure. 

The  volume  of  a  polyhedron  is  thus  the  space  enclosed  by 
its  faces. 

If  two  closed  surfaces  are  placed  side  by  side  so  as  to  have 
some  portion  in  common,  and  this  common  portion  is  disre- 
garded, the  volume  of  the  resulting  figure  is  called  the  sum  of 
the  volumes  of  the  two  given  figures. 

486.  Axiom  13.  If  two  figures  are  identically  equal,  their 
volumes  are  equal. 

487.  Axiom  14.  If  equal  volumes  he  added  to  equal  volumes, 
or  to  the  same  volume,  their  sums  are  equal. 


316 


ELEMENTARY  GEOMETRY 


[Chap.  VII 


Proposition  IV 

488.  An  oblique  prism  is  equal  in  volume  to  a  right 
prism  whose  base  is  a  right  section  of  the  oblique  prism 
and  whose  altitude  is  equal  to  a  lateral  edge  of  the 
oblique  prism. 


Let  AJhe  any  oblique  prism  of  which  F'G'H'J'K'  is  a  right 
section  made  by  the  plane  L. 

It  is  required  to  prove  that  the  prism  AJ  is  equal  in  volume 
to  a  right  prism  whose  base  is  F'G'H'J'K',  and  whose  altitude 
is  equal  to  an  edge  of  AJ. 

Proof.    Produce  the  edge  FA  to  A'  making  F'A'  equal  to  FA. 

Through  A'  pass  a  plane  M  parallel  to  the  plane  L,  and 
meeting  the  other  lateral  edges  of  the  prism  produced  in 
B',  C,  D',  E'. 

Then  the  figure  A' J'  is  a  prism  whose  lateral  edges  are 
equal  to  the  lateral  edges  of  AJ  and  whose  bases  are  right 
sections  of  AJ. 

Does  A' J'  satisfy  the  definition  of  a  prism  ? 

The  quadrilateral  ABB' A'  is  identically  equal  to  the  quadri- 
lateral i^(?  6? 'i^'.     Why? 


488-492]  PBISMS  AND  PYRAMIDS  317 

Similarly,  the  quadrilateral  BCC'B'  is  identically  equal  to 
the  quadrilateral  GHH'G'. 

And  the  polygon  ABODE  is  identically  equal  to  the  poly- 
gon FGHJK,  by  hypothesis. 

Therefore  the  truncated  prism  A'D  is  identically  equal  to 
the  truncated  prism  F'J,  having  the  three  faces  forming  the 
trihedral  angle  B  identically  equal,  two  and  two,  to  the  three 
faces  forming  the  trihedral  angle  G.  (Art.  484.) 

To  each  of  these  truncated  prisms  add  the  figure  AJ'. 

Then  the  volume  of  the  right  prism  ji'J'  is  equal  to  the 
volume  of  the  given  prism  AJ.  (Axioms  13  and  14.) 

Therefore  •  •  • 

Definitions 

489.  A  prism  whose  bases  are  parallelograms  is  called  a 
parallelepiped. 

A  parallelepiped  is  thus  a  hexahedron,  all  of  whose  faces  are 
parallelograms. 

Lateral  faces  which  contain  opposite  sides  of  the  bases  are 
called  opposite  faces. 

490.  A  right  parallelepiped  is  one  in  which  one  set  of  lateral 
edges  is  perpendicular  to  the  bases. 

491.  A  rectangular  parallelepiped  is  a  right  parallelepiped 
whose  bases  are  rectangles. 

492.  A  cube  is  a  parallelepiped  whose  faces  are  all  squares. 


EXERCISES 

1.  At  least  four  of  the  faces  of  a  right  parallepiped  are  rectangles. 

2.  All  the  faces  of  a  rectangular  parallelepiped  are  rectangles. 

3.  If  the  three  faces  of  a  parallelepiped  which  meet  in  any  vertex  are 
squares,  the  figure  is  a  cube. 


318 


ELEMENTARY  GEOMETRY 


[Chap.  VII 


Proposition  V 

493.   The  opposite  lateral  faces  of  any  parallelepiped 
are  parallel  and  identically  equal. 


Let  AG  hQ  2i  parallelepiped  whose  bases  are  ABCD  and 
EFOH. 

It  is  required  to  prove  that  the  opposite  lateral  faces  AF  and 
DG,  also  AH  and  BG  are  parallel  and  identically  equal. 

Proof.  First,  since  AB  is  parallel  to  DC,  and  AE  to  DH, 
the  lateral  face  AF  is  parallel  to  DG.  (Art.  424.) 

Similarly,  AH  is  parallel  to  BG. 

Next,  in  the  parallelograms  AF  and  DG,  AE=DH,  AB=DC, 
and  Z  EAB  =  Z  HDC.  (Art.  424.) 

Therefore  AF  is  identically  equal  to  DG.  (Art.  129.) 

494.  Corollary.  Any  two  opposite  faces  of  a  parallelepiped 
may  he  takeri  as  the  bases. 

When  any  pair  of  opposite  faces  have  been  chosen  as  bases, 
the  edges  meeting  them  must  be  looked  upon  as  the  lateral 
edges. 

EXERCISES 

1.  Any  section  of  a  parallelepiped  cutting  four  edges  between  the 
vertices  is  a  parallelogram. 

2.  Prove  that  in  a  rectangular  parallelepiped  each  edge  is  perpendicular 
to  all  the  edges  meeting  it. 


493-495] 


PRISMS  AND  PYRAMIDS 


319 


Proposition  VI 

495.  The  plane  passed  through  two  diagonally  oppo- 
site edges  of  a  parallelepiped  divides  it  into  two  trian- 
gular prisms  which  are  equal  in  volume. 


Let  BH  be  any  parallelepiped,  and  let  a  plane  be  passed 
through  the  opposite  edges  AE  and  GG, 

It  is  required  to  prove  that  the  triangular  prism  ABC-F  is 
equal  in  volume  to  the  triangular  prism  ADC-H. 

Proof.  Take  a  right  section  PQRS  of  the  parallelepiped,  its 
plane  intersecting  the  plane  through  the  edges  in  the  line  PR. 

Then  PQRS  is  a  parallelogram  of  which  PR  is  a  diagonal. 

The  volume  of  ABC-F  is  equal  to  the  volume  of  a  right  prism 
having  A  PQR  for  base  and  AE  for  lateral  edge.      (Prop.  IV.) 

Similarly,  the  volume  of  ADC-H  is  equal  to  the  volume  of  a 
right  prism  having  A  PSR  for  base  and  AE  for  lateral  edge. 

But  A  PQR  is  identically  equal  to  A  PSR.  (Art.  124.) 

Hence  the  right  prism  whose  base  is  A  PQR  and  lateral  edge 
AE  is  identically  equal  to  the  right  prism  whose  base  is  A  PSR 
and  lateral  edge  AE.  (Art.  482.) 

Therefore  the  triangular  prism  ABC-F  is  equal  in  volume  to 
the  triangular  prism  ADC-H. 


320 


ELEMENTARY  GEOMETRY 


[Chap.  VII 


Proposition  VII 

496.  The  volumes  of  two  rectangular  parallelepipeds 
having  identically  equal  bases  are  in  the  same  ratio  as 
their  altitudes. 


Let  AO  and  AG^  be  two  rectangular  parallelepipeds  having 
their  bases  ABCD  and  A'B'C'D'  identically  equal. 

It  is  required  to  prove  that  the  volume  of  AG  is  to  the  volume 
of  A'G'  as  the  altitude  AE  is  to  the  altitude  A'E'. 

Proof.     Case  I.     When  AE  and  A'E'  are  commensurable. 

Choose  unit  length  which  will  measure  both  AE  and  A'E'. 

Apply  this  unit  length  to  the  altitudes,  and  suppose  that  it 
divides  AE  m  times,  and  A'E'  n  times. 

Then  AE  :  A'E'  =:m:n.  (Art.  220.) 

Through  all  the  points  of  division  of  AE  and  A'E'  pass 
planes  parallel  to  the  bases. 

These  planes  divide  AG  into  m  and  A'G'  into  n  right  paral- 
lelepipeds, which  are  identically  equal.  (Art.  482.) 

volume  of  AG      m 


Hence 


Therefore 


volume  of  A'G'      n 
volume  oi  AG  _  AE 
volume  of  A'G'  ~  A'E' 


496-498]  PRISMS  AND  PYRAMIDS  321 

Case  II.     When  AE  and  A'E'  are  incommensurable. 

Choose  any  unit  length  which  will  measure  AE,  and  apply 
it  as  often  as  possible  to  A'E'. 

Let  P'  be  the  last  point  of  exact  division  so  that  A'P'  is  a 
multiple  of  the  unit  length,  and  the  remainder  P'E'  is  less  than 
the  unit  length. 

Through  P'  pass  a  plane  P'Q'R'S'  parallel  to  the  base. 

Then  A'R'  is  a  parallelepiped,  and  since  AE  and  A'P'  are 
commensurable, 

volume  of  AG  ^  AE  CCase  I  ^ 

volume  of  ^'i?'      A'P'  ^  '^ 

If  now  the  length  of  the  unit  measure  be  repeatedly  di- 
minished, the  remainder  P'E'  can  be  made  smaller  than  any 
assigned  quantity,  so  that  ^'P'  approaches  A'E'  as  its  limit, 
and  the  parallelepiped  A'B'  approaches  the  parallelepiped  A'G' 
as  its  limit. 

Also,    ^0'"'"^  »f  ^Cf    ^in  approach    ^°'"'""  "^  "^^ 
volume  of  A'W  volume  of  A'G' 

as  its  limit. 

And  will  approach  — — -  as  its  limit. 

-D  ,  volume  of  AG  _  AE 

volume  QiAR'~~^^' 
for  every  unit  of  measure. 

Therefore  volume  of  ^(^^  ^  AE^^ 

volumeof  ^'(?'      AE'  ^    it.  ^ou.; 

497.  Definition.  The  lengths  of  the  three  edges  which 
meet  in  any  vertex  are  called  the  dimensions  of  a  rectangular 
parallelepiped. 

498.  Theorem.  If  two  rectangular  parallelepipeds  have  two 
dimensions  of  one  equal,  respectively,  to  two  dimensions  of  the  other, 
their  volumes  are  in  the  same  ratio  as  their  third  dimensions. 

This  is  merely  a  restatement  of  Proposition  VII. 

Y 


322 


ELEMENTARY  GEOMETRY 


[Chap.  VH 


Proposition  VIII 

499.  The  volumes  of  two  rectangular  parallelepipeds 
having  equal  altitudes  are  in  the  same  ratio  as  the 
areas  of  their  bases. 


\ 


\ 

Q 

|\ 

c 

k 

-A 

h' 


b' 


Let  P  and  Q  be  the  volumes  of  two  rectangular  parallele- 
pipeds having  equal  altitudes  c,  and  the  dimensions  of  whose 
bases  are  a,  b,  and  a',  b\ 

It  is  required  to  prove  that 

P      a  xb 


Q     a'xb' 

Proof.     Construct  a  third  rectangular 

parallelepiped  whose 

volume  is  R,  and  whose  dimensions  are  a,  b',  c. 

Then 

P     b 
R     b" 

(Art.  498.) 

also 

R      a 
Q     a'' 

Therefore 

r""  Q-b'^'a" 

or 

P      axb 
Q     a'  xb' 

500.  Theorem.  If  tico  rectangular  parallelepipeds  have  one 
dimension  of  the  one  equal  to  one  dimeyision  of  the  other,  their 
volumes  are  in  the  same  ratio  as  the  products  of  their  other  tivo 
dimensions. 


This  is  again  a  restatement  of  Proposition  VIII. 


499-502] 


PRISMS  AND  PYRAMIDS 


323 


Proposition  IX 

501  The  volumes  of  any  two  rectangular  parallele- 
pipeds are  in  the  same  ratio  as  the  products  of  their 
three  dimensions. 


Let  P  and  Q  be  the  volumes  of  two  rectangular  parallele- 
pipeds whose  dimensions  are  a,  6,  c,  and  a',  6',  c',  respectively. 
It  is  required  to  prove  that 

P  _  a  X  b  X  c 
Q     a'  xb'  X  c' 

Proof.     Construct  a  third  rectangular  parallelepiped  whose 
volume  is  B  and  whose  dimensions  are  a',  b',  c. 


Then 


and 


Therefore 


or 


P 
R 

E 
Q 

^x? 

P 
Q 


a  xb 
a'  X  b'' 

c 
c'' 

a  xb       c 

a'  X  b'     c' 

a  X  b  X  c 
a'  xb'  X  c' 


(Art.  500.) 
(Art.  498.) 


502.  In  order  to  express  the  measure  of  any  volume  it  is 
first  necessary  to  select  a  unit  volume,  with  which  the  given 
volume  is  to  be  compared. 

The  unit  volume  usually  chosen  is  the  volume  of  a  cube, 
each  of  whose  edges  is  of  unit  length. 


324 


ELEMENTARY  GEOMETRY 


[Chap.  VII 


When  the  unit  of  length  is  a  foot,  the  unit  of  volume  is  the 
volume  of  a  cube  each  of  whose  edges  is  a  foot  in  length ;  or, 
as  we  say,  the  unit  of  volume  is  a  cubic  foot. 

When  the  unit  of  length  is  an  inch,  or  a 
yard,  the  unit  of  volume  is  a  cubic  inch, 
or  a  cubic  yard. 

The  measure  of  a  volume  is  the  number 
which  expresses  how  many  times  it  will 
contain  the  unit  of  volume. 

If  in  a  rectangular  parallelepiped,  the 
three  edges  which  meet  in  any  vertex  are  commensurable 
magnitudes,  and  these  are  divided  into  unit  lengths,  the  whole 
volume  can  be  divided  into  unit  volumes  by  passing  planes 
through  the  points  of  division.  Thus  the  parallelepiped  in  the 
diagram  contains  thirty-six  units  of  volume. 

For  brevity  we  frequently  say  "the  volume  of  a  parallele- 
piped'' is  a  certain  number,  when  it  would  be  more  exact  to 
say  "  the  measure  of  the  volume." 

In  the  preceding  proposition,  if  the  parallelepiped  whose 
volume  is  Q  should  have  each  edge  of   unit  length,  so  that 

a'  =  6'  =  c'  =  l, 


then  Q  would  be  the  unit  volume. 
In  that  case 


Q 


a  xb  X  c, 


or 


P=a  X  h  X  c  times  unit  volume. 


That  is,  a  X  6  X  c  is  the  measure  of  the  volume  P. 

503.   Theorem.     The  measure  of  the  volume  of  any  rectangu- 
lar parallelepiped  is  the  product  of  its  three  dimensions. 


504.   Corollary.     TJie  volume  of  ayiy  rectangular  parallele- 
nped  is  the  product  of  its  altitude  and  the  area  of  its  base. 


502-505] 


PBISMS  AND  PYRAMIDS 


325 


Proposition  X 

505.   The  volume  of  any  parallelepiped  is  equal  to  the 
product  of  its  altitude  and  the  area  of  its  hase. 


E 


,/\^" 

/■ 

1  / 

F 

/d" 

/ 

Let  ^(t  be  any  oblique  parallelepiped  whose  base  is  ABCD 
and  altitude  KE. 

It  is  required  to  prove  that  the  volume  of  ^6r  is  equal  to  the 
product  of  KE  and  the  area  of  ABCD. 

Proof.  Produce  the  edges  AB,  DC,  HG,  EF,  and  cut  them 
perpendicularly  at  A',  D',  H',  E',  and  B,'  C,'  G',  F',  by  two 
parallel  planes  whose  distance  apart  A'B'  is  equal  to  AB. 

Then  A'G'  is  a  right  parallelepiped,  A'H'  being  the  base, 
equal  in  volume  to  AG.  (Prop.  IV.) 

Again,  produce  the  edges  D'A',  C'B',  G'F',  H'E',  of  the 
parallelepiped  A'G',  and  cut  them  perpendicularly  at  D",  C", 
G",  H",  and  A",  B",  F",  E",  by  two  parallel  planes  whose  dis- 
tance apart  B"C"  is  equal  to  B'C. 

Then  A"G"  is  a  rectangular  parallelepiped  [why?]  equal  in 
volume  to  A'G'  (Prop.  IV),  and  hence  equal  to  AG. 

Now  the  volume  of  A"G"  is  equal  to  the  product  of  the  alti- 


tude A"E"  and  the  area  of  the  base  A"C' 


(Art.  504.) 


But  since  all  three  parallelepipeds  lie  between  the  same  two 
parallel  planes,  the  altitude  KE  equals  the  altitude  A"E". 


326 


ELEMENTARY  GEOMETRY 


[Chap.  VII 


Also  the  parallelograms  ACj  A'C,  and  A"C"  are  equal  in 
area.  (Art.  294) 

Therefore,  volume  of  ^G^  =  volume  of  A"G" 

=  area  of  A"C"  x  A"E'' 
=  area  of  AG  x  KE. 


Proposition  XI 

506.   The  volume  of  a  triangular  prism  is  equal  to 
the  product  of  its  altitude  and  the  area  of  its  base. 


Let  ABC-DEF  be  any  triangular  prism  whose  altitude  is  h. 

It  is  required  to  prove  that  the  volume  of  this  prism  is  equal 
to  h  times  the  area  of  the  base  ABC. 

Proof.  Complete  the  parallelogram  ^jBC^.  Through /f  draw 
a  line  parallel  to  a  lateral  edge  AD,  meeting  the  plane  DEF 
at  L,  and  complete  the  parallelepiped  BL  having  the  same 
altitude  as  the  prism. 

The  prism  ABC-DEF  is  one-half  of  the  parallelepiped 
BL.  (Prop.  VI.) 

Now  the  volume  of  BL  is  the  prodixit  of  its  altitude  h  and 
the  area  of  the  base  ABGK.  (Prop.  X.) 

Therefore  the  volume  of  the  prism  ABG-DEF  is  equal  to 
h  times  the  area  of  ABG^  which  is  half  of  the  base  ABGK. 


505-508] 


PRISMS  AND  PYRAMIDS 


327 


507.  Corollary  I.  Hie  volume  of  any  prism 
is  equal  to  the  product  of  its  altitude  and  the  area 
of  its  base. 

Any  prism  can  be  divided  into  triangular  prisms  by 
passing  planes  through  one  lateral  edge  and  all  the  non- 
adjacent  lateral  edges. 

508.  Corollary  II.  The  volumes  of  any  two 
prisms  are  equal  if  they  have  equal  altitudes  and 
bases  of  equal  areas. 


,  --'-  ~ 


EXERCISES 

1.  The  square  on  the  diagonal  of  a  rectangular  parallelepiped  is  equal 
to  the  sum  of  the  squares  on  the  three  edges  meeting  in  one  extremity  of 
the  diagonal. 

2.  The  square  on  the  diagonal  of  a  cube  is  three  times  the  square  on 
one  of  its  edges. 

3.  Show  that  all  the  diagonals  of  a  rectangular  parallelepiped ^are 
equal. 

4.  The  sum  of  the  squares  on  the  four  diagonals  of  a  parallelepiped  is 
equal  to  the  sum  of  the  squares  on  the  twelve  edges. 

5.  Every  section  of  a  prism  made  by  a  plane  parallel  to  an  edge  is  a 
parallelogram. 

6.  Find  the  volume  and  the  length  of  the  diagonal  of  a  cube  whose 
edge  is  3  feet. 

7.  Show  that  all  the  diagonals  of  any  parallelepiped  pass  through  one 
point,  and  are  bisected  at  that  point. 

8.  The  edge  of  a  cube  is  4  feet.  Find  the  edge  of  a  cube  having  twice 
the  volume. 

9.  The  dimensions  of  a  rectangular  parallelepiped  are  3,  5,  and  7 
decimeters.    Find  its  volume  and  the  length  of  its  diagonals. 

10.  A  box,  covered  top  and  bottom,  which  is  made  of  boards  2  inches 
thick,  has  outside  dimensions  of  18,  24,  and  30  inches.  Find  its  entire 
contents. 

11.  Find  the  volume  of  a  right  prism  whose  base  is  a  regular  hexagon 
of  6  inches  side,  and  whose  altitude  is  15  inches. 

12.  The  diagonal  of  a  cube  is  27  inches,  find  its  volume. 


328  ELEMENTARY   GEOMETRY  [Chap.  VII 

Section  II 

PYRAMIDS 

509.  A  pyramid  is  a  polyhedron  one  of  whose  faces  is  a  poly- 
gon, and  the  remaining  faces  are  triangles  having  a  common 
vertex. 

The   polygonal   face   is   called  the  base  of  the 
pyramid,  the  triangular  faces  are  the  lateral  faces, 
and  the  common  vertex  of  the  lateral  faces  is  the      / 
vertex  of  the  pyramid.  A' 

The  sum  of  the  areas  of  the  lateral  faces  is  called  ^ 
the  lateral  area  of  the  pyramid. 

The  perpendicular  distance  from  the  vertex  to  the  base  is 
the  altitude  of  the  pyramid. 

A  triangular  pyramid  is  one  whose  base  is  a  triangle  •,  a  quad- 
rangular pyramid,  one  whose  base  is  a  quadrilateral,  etc. 

A  triangular  pyramid  is  a  tetrahedron  since  it  has  four  faces. 

All  the  faces  of  a  tetrahedron  are  triangles.  Any  face  may 
be  taken  as  the  base,  and  the  opposite  vertex  as  the  vertex. 

How  many  vertices  has  a  tetrahedron  ?  how  many  edges  ?  how 
many  pairs  of  non-intersecting  edges  ?  These  are  called  the 
pairs  of  opposite  edges. 

510.  A  regular  pyramid  is  one  whose  base  is  a  regular  poly- 
gon and  whose  vertex  lies  on  the  perpendicular  to  the  base 
drawn  from  its  centre. 

The  slant  height  of  a  regular  pyramid  is  the  altitude  of  any 
one  of  its  lateral  faces;  i.e.  it  is  the  perpendicular  distance 
of  the  vertex  from  any  side  of  the  base. 

511.  A  truncated  pyramid  is  the  figure  formed  by  the  base 
and  any  section  of  a  pyramid,  and  the  por-  ^r-—-""^^^ 
tions  of  the  lateral  faces  intercepted  between         A     I    /\ 
them.                                                                           /  /--!--. A   \ 

If  the  upper  section  is  parallel  to  the  base,  V  /  \  / 
the  figure  is  called  a  frustum  of  a  pyramid.  ^ ^ 


509-512]  PRISMS  AND  PYRAMIDS  329 

The  altitude  of  a  frustum  of  a  pyramid  is  the  perpendicular 
distance  between  its  bases. 

The  slant  height  of  the  frustum  of  a  regular  pyramid  is  the 
perpendicular  distance  between  the  parallel  edges  of  a  lateral 
face. 

Proposition  XII 

512.  The  lateral  area  of  a  regular  pyramid  is  equal 
to  one-half  the  product  of  the  perimeter  of  the  base  and 
the  slant  height. 


Let  S- ABODE  be  any  regular  pyramid  whose  slant  height 
isASP. 

It  is  required  to  prove  that  its  lateral  area  is  equal  to  one- 
half  the  product  of  the  perimeter  of  the  polygon  ABGDE  and 
the  slant  height  SP. 

Proof.     1.    What  is  the  area  of  A  SAB  ?  (Art.  306.) 

2.  As  SAB,  SBC,  etc.,  have  equal  altitudes. 

3.  What  is  the  area  of  all  the  triangles  ? 

4.  What  is  the  lateral  area  ? 

EXERCISES 

1.  Prove  that  the  lateral  edges  of  a  regular  pyramid  are  all  equal,  and 
hence  that  the  lateral  faces  are  equal  isosceles  triangles. 

2.  Show  that  the  lateral  faces  of  the  frustum  of  a  pyramid  are  all 
trapezoids,  and  of  a  regular  pyramid  are  all  equal  trapezoids. 

3.  Show  that  the  lateral  area  of  the  frustum  of  a  regular  pyramid  is 
equal  to  half  the  sum  of  the  perimeters  of  its  bases  multiplied  by  the 
slant  height. 


330  ELEMENTARY  GEOMETRY  [Chap.  VII 


Proposition  XIII 

513.  If  a  pyramid  is  cut  by  a  plane  parallel  to  its 
base,  the  edges  and  the  altitude  are  divided  in  the  same 
ratio,  the  section  made  by  the  plane  is  sunilar  to  the 
base,  and  the  area  of  the  section  is  to  the  area  of  the 
base  as  the  squares  of  their  distances  from  the  vertex. 


Let  S-ABCDE  be  any  pyramid  whose  altitude  is  SO,  and 
let  it  be  cut  by  a  plane  L  parallel  to  the  base,  in  the  polygon 
A'B'C'D'E'  whose  distance  from  S  is  SO'. 

It  is  required  to  prove 

1.  That  M  =  ^  =  ^==etc.  =  ^. 

SA      SB      SO  SO 

2.  That  A'B'C'D'E'  is  similar  to  ABODE.  _ 

3.  That  area  of  A'B'C'D'E' :  area  of  ABODE = SO^ '•  SO^- 

Proof.     First.    In  A  SAB,  A'B'  is  parallel  to  AB.    (Art.  419.) 

Therefore  —  =  ^'.  (Art.  242.) 

SA      SB  ^  ^ 

Q.    -1    1  SB'     SC 

Similarly,  ^  =  ^^  *  '  ' 

Also,  since  0'  must  lie  upon  SO,  A'O'  is  parallel  to  AO. 

Therefore  SA^^SO^^  (Art.  242.) 

SA      SO  ^  ^ 

Whence  M  =  ^  =  ^=.  ..  =  ^. 

SA      SB      SC  SO 


AB      SB 

SB'     B'C 

SB       BC' 

A'B' 

B'C      CD' 

513-516]  PRISMS  AND  PYRAMIDS  331 

Second.     Since  A'B'  is  parallel  to  AB,  and  B'C  is  parallel 

to  BC, 

Z  A' B'C  =  Z.ABa  (Art.  424.) 

Similarly,  Z  B'CD'  =  Z  BCD,  .  .  . 

Hence  the  polygons  A'B'CD'E'  and  ABODE  are  mutually 
equiangular. 

.,  A'B'     SB' 

Also, 

'    Similarly, 

Therefore  ^=^  =  ^  =  «t<'- 

Hence,  in  the  polygons  A'B'CD'E'  and  ABODE,  pairs  of 
homologous  sides  are  in  the  same  ratio. 

Therefore  the  polygons  A'B'O'D'E'  and  ABODE  are  similar. 

Third.     Area  of  A'B'O'D'E' :  area  of  ABODE 

=  AJB^'  :  AS  (Art.  315.) 

=  SA^ :  SAL 

=  SO'' :  so". 

514.  Corollary  I.  Sections  of  the  same  pyramid  made  by 
parallel  planes  are  similar. 

515.  Corollary  II.  Hie  areas  of  parallel  sections  of  a 
pyramid  are  in  the  same  ratio  as  the  squares  of  their  distances 
from  the  vertex. 

516.  Corollary  III.  If  two  2W^^^^^  ^^^^^  equal  altitudes 
and  bases  of  equal  areas,  the  areas  of  sections  made  by  planes 
equidistant  from  their  vertices  are  equal. 


332 


ELEMEN TARY  GEOMETR Y 


[Chap.  Vll 


Proposition  XIV 

517.   If  two  triangular  pyramids  have  equal  altitudes 
and  equal  bases,  they  have  equal  volumes. 


Let  S-ABC  and  S'-A'B'C  be  two  triangular  pyramids  hav- 
ing their  bases  ABC  and  A'B'C  equal  in  area,  and  their  alti- 
tudes SO  and  S'O'  also  equal. 

It  is  required  to  prove  that  the  volume  of  S-ABC  is  equal  to 
the  volume  of  S'-A'B'C. 

Proof.  Divide  the  altitudes  of  the  pyramids  S  and  S'  into 
any  number  of  equal  parts,  each  into  the  same  number,  and 
through  the  points  of  division  pass  planes  parallel  to  the  bases. 

Let  the  sections  made  by  these  planes  be  DEF,  GHK,  etc., 
in  the  one  pyramid,  and  D'E'F',  G'H'K',  etc.,  in  the  other. 

Then 

the  area  of  DEF  =  the  area  of  D'E'F',       (Art.  516.) 

area  of  OHK=  area  of  G'H'K',  etc. 

With  the  sections  DEF,  GHK,  etc.,  as  upper  bases,  con- 
struct prisms  Y,  Z,  etc.,  having  their  edges  equal  and  parallel 
to  DA,  GD,  etc. 

Also  with  A'B'C,  D'E'F',  G'H'K',  etc.,  as  lower  bases,  con- 
struct prisms  X',  Y',  Z',  etc.,  having  their  edges  equal  and 
parallel  to  D'A',  G'D',  S'G',  etc. 


517-520]  PRISMS  AND  PYRAMIDS  333 

These  prisms  are  all  of  the  same  altitude,  and  in  volume 
T=  Y',  Z=  Z\  etc.  (Art.  508.) 

Therefore  the  sum  of  all  the  prisms  in  the  figure  S^  is  greater 
than  the  sum  of  all  the  prisms  in  the  figure  8  by  the  lowest 
prism  X'. 

Denote  the  sum  of  the  volumes  of  the  prisms  in  the  figure  S 
by  P,  in  the  figure  >S'  by  P,  and  the  volume  of  X '  by  E. 

Then  P=P'-E. 

Now,  if  the  number  of  parts  into  which  the  altitudes  are 
divided  is  indefinitely  increased,  and  the  altitudes  of  the  prisms 
consequently  indefinitely  diminished,  P  will  approach  the 
volume  of  the  pyramid  S-ABC  as  its  limit,  P'  will  approach 
the  volume  of  the  pyramid  S'-A'B'C  as  its  limit,  while  E  will 
become  indefinitely  small,  and  is  ultimately  less  than  any 
assignable  quantity. 

Therefore  the  difference  between  the  volumes  of  the  pyra- 
mids S-ABC  and  S'-A'B'C  is  less  than  any  assignable  quantity. 

That  is,  the  volumes  of  the  two  pyramids  are  equal. 

518.  Note.     The  test  for  geometrical  equality  is  superposition. 
Tliat  is  to  say,  two  geometrical  magnitudes  are  equal  when  they  can  be 

made  to  occupy  the  same  position. 

The  test  for  numerical  equality  is  given  by  the  following  definition  : 
Two  numerical  magnitudes  are  equal  when  their  difference  is  less  than 

any  assignable  quantity. 

519.  Corollary  I.  Triang2ilar pyramids  upon  the  same  base 
and  having  their  vertices  in  a  plane  parallel  to  the  base,  are  equal 

•  in  volume. 

The  vertex  of  a  tetrahedron  may  be  moved  anywhere  in   a 
,     plane  parallel  to  the  base  without  altering  the  volume. 

520.  Corollary  II.  Triangular  pyramids  having  equal 
bases  in  the  same  plane  and  their  vertices  at  the  same  point,  are 
equal  in  volume. 

If  the  base  of  a  pyramid  is  a  parallelogram,  the  plane  through 
the  vertex  and  two  opposite  vertices  of  the  base  divides  the 
pyramid  into  two  triangular  pyramids  having  equal  volumes. 


334  ELEMENTARY  GEOMETRY  [Chap.  VII 


Proposition  XV 

521.  The  volume  of  a  triangular  pyramid  is  one- 
third  the  volum^e  of  a  triangular  j)risT)%  of  the  same 
base  and  altitude. 


Let  S-ABC  be  any  triangular  pyramid  and  ABC-STV  be  a 
triangular  prism  having  the  same  base  and  altitude. 

It  is  required  to  prove  that  the  volume  of  S-ABC  is  equal  to 
one-third  the  volume  of  ABC-STV. 

Proof.  In  the  face  BCVT  of  the  prism  draw  the  diagonal 
CTj  and  through  the  points  S,  C,  T,  pass  a  plane. 

The  given  prism  is  thus  composed  of  the  three  triangular 
pyramids  S-ABC,  S-BCT,  and  S-VCT. 

Now  the  volumes  of  the  two  pyramids  S-BCT  and  S-VCT 
are  equal.     Why  ? 

The  pyramid  S-VCT  may  also  be  read  C-SVT 

And  the  volume  of  C-S  VT  is  equal  to  the  volume  of  S-ABC. 
Why? 

Therefore  the  prism  ABC-STV  is  composed  of  three  tri- 
angular pyramids,  each  of  which  is  equal  in  volume  to  S-ABC. 

Therefore  the  volume  of  S-ABC  is  equal  to  one-third  of  the 
volume  of  the  prism  ABC-STV 

522.  Corollary  I.  The  volume  of  a  triangular  pyramid  is 
equal  to  one-third  the  product  of  its  altitude  and  the  area  of  its 
base.  (Art.  506.) 


521-525]  PRISMS  AND  PYRAMIDS  335 

523.  Corollary  II.  The  volume  of  any  pyramid  is  equal  to 
one-third  the  product  of  its  altitude  and  the  area  of  its  base. 

The  pyramid  can  be  divided  into  triangular  pyramids    by 
passing  planes  through  the  vertex. 

524.  Corollary  III.  TJie  volumes  of  any  two  pyramids  are 
in  the  same  ratio  as  the  products  of  their  altitudes  by  the  areas 
of  their  bases. 

If  V  and  V  are  the  volumes  of  two  pyramids,  A  and  A'  the 
areas  of  their  bases,  h  and  h'  their  altitudes, 

then  V=^hA,     V  =  i  h'A'.  (Art.  523.) 

Whence  i:  =  lM.=  hA, 

V     \h<A>     h'A' 


525.  Corollary  IV.  Tlie  volumes  of  two  pyramids  having 
equal  altitudes  are  in  the  same  ratio  as  the  areas  of  their  bases, 
or  having  equal  bases  are  in  the  same  ratio  as  their  altitudes. 

EXERCISES 

1.  A  pyramid  is  cut  by  a  plane  midway  between  the  vertex  and  the 
base.    Find  the  ratio  of  the  area  of  the  section  to  the  area  of  the  base. 

2.  In  the  figure  of  Proposition  XIV,  assuming  the  altitude  of  the  pyr- 
amid S  to  have  been  divided  into  three  equal  parts,  show  that  the  volume 
of  the  prism  Y  is  four  times  the  volume  of  the  prism  Z. 

3.  The  section  of  a  pyramid  made  by  a  plane  parallel  to  the  base  is 
equal  in  area  to  four-ninths  of  the  area  of  the  base.  In  what  ratio  does 
the  plane  of  section  divide  the  lateral  area  ? 

4.  A  regular  pyramid  and  a  regular  prism  have  equal  hexagonal  bases, 
and  altitudes  equal  to  three  times  the  radius  of  the  circles  circumscribing 
the  bases.     What  is  the  ratio  of  their  lateral  areas  ? 

6.  Find  the  lateral  area  and  the  volume  of  a  regular  octagonal  pyramid 
whose  slant  height  is  5  metres  and  a  side  of  whose  base  is  2  metres. 

6.  The  total  surface  of  a  regular  pyramid,  whose  base  is  a  square  of  10 
feet  side,  is  360  square  feet.    Find  its  altitude. 


336  ELEMENTARY  GEOMETRY  [Chap.  VII 


Proposition  XVI 

526.  The  volumes  of  two  tetrahedrons  having  a  tri- 
hedral angle  of  the  one  equal  to  a  trihedral  angle  of  the 
other,  are  in  the  same  ratio  as  the  products  of  the  edges 
which  meet  in  the  vertices  of  these  angles. 


G' 
Let  0-ABC  and  O-A'B'C  be  two  tetrahedrons  having  the 
trihedral  angles  at  0  equal,  and  let  their  volumes  be  Fand  V. 
It  is  required  to  prove  that 

V:  V'  =  A0xB0xC0:  A'O  x  B'O  x  CO. 

Proof.  Place  the  tetrahedrons  so  that  their  equal  trihedral 
angles  coincide. 

From  A  and  A'  draw  perpendiculars  AP  and  A'P'  to  the 
plane  OBO.  These  will  lie  in  a  plane  with  OAA'  and  will 
meet  OBC  in  points  of  a  straight  line  OPP'.  (Art.  405.) 

Then  A^  =  ^.     Why? 

A'P'     A'O  ^ 


Also 


volume  of  0-ABC 
volume  of  0-A'B'C 

^    APx  area  of  OBC 
A'P'  X  area  of  OB'C 
AO      area  of  OBC 


(Art.  524.) 


A'O    area  of  OB'C 

Now  area  of  0^(7^^0x00.         (Art.  308.) 

area  of  O^'O'     B'O  x  CO  ^  ^ 

Therefore    ^^^^™®  ^^  0-ABC       AO  x  BO  x  CO 


volume  of  0-A'B'C     A'O  x  B'O  x  CO 


526-527]  PRISMS  AND  PYRAMIDS  337 


Proposition  XVII 

527.  The  volume  of  a  truncated  triangular  prism  is 
equal  to  the  sum  of  the  volumes  of  three  pyramids  whose 
common  base  is  one  base  of  the  prism  and  whose  vertices 
are  the  three  vertices  of  the  other  base. 


Let  ABC-DEF  be  any  truncated  triangular  prism. 

It  is  required  to  prove  that  the  volume  of  this  figure  is 
equal  to  the  sum  of  the  volumes  of  three  pyramids  whose 
common  base  is  ABC,  and  whose  vertices  are  D,  E,  F,  respec- 
tively. 

Proof.  Through  the  points  E,  A,  C,  and  E,  D,  C,  pass 
planes,  thus  dividing  the  given  figure  into  the  three  pyramids 
E-ABC,  E-ADC,  and  E-DFC. 

The  first  of  these,  viz.  E-ABC,  has  the  base  ABC  and  ver- 
tex jEJ. 

The  second,  E-ABC  may  be  read  C-AED. 

Now  the  base  AED  is  equal  in  area  to  the  base  ABD.  (Art.  296.) 

Therefore  the  volume  of  C-AED  is  equal  to  the  volume  of 
C-ABD  (Prop.  XIV),  which  is  the  same  as  D-ABC. 

The  third,  E-DFC,  is  equal  in  volume  to  E-AFC,  which  is 
the  same  as  A-EFC.  And  this  is  equal  in  volume  to  A-BFC, 
which  is  the  same  as  F-ABC 

Therefore  the  prism  ABC-DEF  is  equal  in  volume  to  the 
pyramids  E-ABC  ^- D-ABC  ^- F-ABC. 


338 


ELEMENTARY  GEOMETRY 


[Chap.  VII 


528.  Corollary  I.  The  volume  of  a  truncated  right  tri- 
angular prism  is  equal  to  the  product  of  the  ai-ea  of  that  face  to 
which  the  edges  are  perpendicular  and  one-third 
the  sum  of  the  lateral  edges. 

For  the  lateral  edges  are  the  altitudes  of  the 
three  pyramids  to  the  sum  of  which  the 
volume  of  the  prism  is  equal. 


529.  Corollary  II.  The  volume  of  any  trun- 
cated triangular  prism  is  equal  to  the  product  of 
the  area  of  a  right  section  and  one-third  the  sum 
of  the  lateral  edges. 

On  either  side  of  the  right  section  lies  a  truncated  right  prism 


EXERCISES 

1.  Any  two  vertices  of  a  tetrahedron  may  move  along  straight  lines 
drawn  through  them  parallel  to  the  edge  through  the  other  two,  without 
altering  the  volume  of  the  tetrahedron. 

The  proof  of  this  theorem  is  contained  in  Proposition  XVII, 
where  it  was  shown  that  the  volume  of  the  tetrahedron 
E  —  DEC  is  unaltered  when  E  moves  to  J5,  and  D  to  A. 

2.  If  A  is  the  area  of  the  lower  base  of  a  frustum  of  a  pyramid,  a  the 
area  of  the  upper  base,  and  h  the  altitude,  then  the 

volume  of  the  frustum  =  -{A-\-  a-\-  ^/Aa).  /J\"n^ 

Let  I  be  the  altitude,  above  the  base  a,  of  the  /  |  \  \ 

pyramid  of  which  the  figure  considered  is  a  frustum.  /    j  \     \ 

Then  A:  a  ={h  +  ly  :l\ 

or  y/A  :  Va  =  h  -h  1:1, 

VA-Va:Va  =  h:l;     (Art.  237.) 


and 
whence 

Va  —  Va  Va 

Yolnme  =-A(h-i- I) -I  al 


1  = 


hVA 


^HAVA-aValh^j^^^^^^^^ 
3L    VA-Va  J     3 


VA-Va 

ir    AhVA ahVa     - 

sLv^-Va      VA-Va. 


528-532]  PRISMS  AND  PYRAMIDS  339 

Section  III 

SIMILAR   POLYHEDRONS 

530.  Definition.  Two  polyhedrons  are  said  to  be  similar 
when  they  have  the  same  number  of  faces  similar  each  to  each 
and  similarly  placed,  and  have  their  corresponding  polyhedral 
angles  identically  equal. 

Similar  polyhedrons  not  only  have  the  same  number  of  faces, 
but  also  the  same  number  of  vertices  and  the  same  number  of 
edges.  They  have  the  same  number  of  faces  meeting  at  corre- 
sponding vertices,  and  consequently  the  same  number  of  edges 
meeting  at  corresponding  vertices. 

If  two  prisms  are  similar,  their  bases  must  be  similar  polygons, 
and  their  homologous  lateral  faces  similar  parallelograms  ; 
while  if  two  pyramids  are  similar,  their  bases  are  similar 
polygons,  and  their  homologous  lateral  faces  are  similar 
triangles. 

531.  Since  corresponding  faces  in  two  similar  polyhedrons 
are  similar  polygons,  and  homologous  sides  of  two  similar  poly- 
gons are  proportional, 

Therefore,  any  two  edges  or  diagonals  of  a  polyhedron  are  in 
the  same  ratio  as  the  homologous  edges  or  diagonals  in  a  similar 
polyhedron. 

Show  that  this  is  true  when  the  edges  do  not  lie  in  the  same 
face  of  the  polyhedron. 

532.  The  areas  of  two  similar  polygons  are  in  the  same 
ratio  as  the  squares  of  any  two  homologous  sides.      (Art.  315.) 

Therefore,  the  surfaces  of  two  similar  polyhedrons  (i.e.  the 
sums  of  the  areas  of  all  the  faces)  are  in  the  same  ratio  as  the 
squares  of  any  two  homologous  edges  or  homologous  diagonals. 

EXERCISES 

1.  A  plane  parallel  to  the  base  cuts  from  the  top  of  a  pyramid  a  figure 
similar  to  the  given  pyramid.     Can  the  same  be  said  of  a  prism  ? 


340  ELEMENTARY  GEOMETRY  [Chap.  VII 


Proposition  XVIII 

533.  Two  tetrahedrons  are  similar  if  the  three  faces 
meeting  in  a  vertex  of  the  one  are  similar  and  similarly 
situated  to  the  three  faces  meeting  in  a  vertex  of  the 
other. 

A 


If  the  tetrahedrons  are  A-BCD  and  A'-B'C'D',  and  the  faces 
meeting  at  the  vertex  A  are  similar  and  similarly  situated  to 
the  faces  meeting  at  the  vertex  A',  the  face  opposite  A  is  simi- 
lar and  similarly  situated  to  the  face  opposite  A',  since  the 
three  sides  of  one  of  these  faces  are  proportional  to  the  three 
sides  of  the  other.  (Art.  253.) 

Also  the  trihedral  angle  A  is  equal  to  the  trihedral  angle  A\ 
since  the  three  face  angles  of  the  one  are  equal,  respectively, 
and  are  similarly  placed  to  the  three  face  angles  of  the  other. 

(Art.  461.) 

For  the  same  reason  the  trihedral  angles  B,  C,  D,  are  equal, 
respectively,  to  the  trihedral  angles  B',  G\  D\ 

Therefore  the  tetrahedrons  are  similar. 

534.  Theorem.  Any  two  similar  polyhedrons  can  he  divided 
into  the  same  number  of  tetrahedrons,  similar  two  and  two. 

For,  if  AB^  AC,  AD,  and  A'B',  A'C,  A'D'  are  consecutive  homologous 
edges  of  the  two  polyhedrons,  meeting  in  the  homologous  vertices  A  and 
A'  the  three  pairs  of  planes  determined  by  these  edges,  together  with  the 
planes  BCD  and  B'C'D',  form  the  faces  of  two  tetrahedrons  which  are 
similar  by  the  above  proposition,  and  are  similarly  placed.  This  process 
of  cutting  off  similar  tetrahedrons  can  be  continued  till  only  two  similar 
tetrahedrons  remain. 


633-536]  PBISMS  AND  PYRAMIDS  341 


Proposition  XIX 

535.   Tlie  volumes  of  two  similar  tetrahedrons  are  in 
the  same  ratio  as  the  cubes  of  their  homologous  edges. 

Let   A-BCD  and   A'-B'C'D'   be  two   similar  tetrahedrons 
whose  volumes  are  denoted  by   Fand  V. 

It  is  required  to  prove  that  V :  V  =  A&  :  J^l 

Proof.     If  A  and  A'  are  two  homologous  vertices,  the  tri- 
hedral angles  A  and  A'  must  be  equal  by  definition. 

n^i,      V  F_    ABy<ACxAD  ,p         ^^. . 

Therefore  ^  -  ^,^^  ^  ^,^,  ^  ^,^,  (Prop.  XVI.) 

AB  ^AC^^  AD 


A'B'     A'C      A'D' 
Therefore  Z.,  =  A^.A^.^B       ^ 


V     A'B'     A'B'     A'B'      JJB^' 

536.  Corollary.  TJie  volumes  of  any  two  similar  polyhe- 
drons are  in  the  same  ratio  as  the  cubes  of  ttvo  homologous  edges. 

For  the  polyhedrons  can  be  divided  into  the  same  number  of 
similar  tetrahedrons,  and  any  two  of  these  which  are  simi- 
larly placed  are  in  the  same  ratio  as  the  cubes  of  homologous 
edges.  Hence  the  sums  of  all  the  tetrahedrons  are  in  the 
same  ratio  as  the  cubes  of  homologous  edges.    (Art.  240.) 

EXERCISES 

1.  Any  two  cubes  are  similar  polyhedrons,  and  hence  their  volumes 
are  in  the  same  ratio  as  the  cubes  of  homologous,  or  any,  edges. 

2.  A  plane  is  passed  through  the  mid-points  of  three  edges  of  a  tetra- 
hedron ;  compare  the  volume  of  the  original  tetrahedron  with  the  volumes 
of  the  parts  into  which  it  is  divided  by  the  plane. 


342  ELEMENTARY  GEOMETRY  [Chap.  VII 

Section  IV 

REGULAR   POLYHEDRONS 

537.  Definition.  A  regular  polyhedron  is  one  whose  faces 
are  equal  regular  polygons  and  whose  polyhedral  angles  are  all 
equal. 

Since  the  polyhedral  angles  are  all  equal,  the  same  number 
of  faces  and  edges  must  meet  at  each  vertex ;  and  since  the 
faces  are  all  equal,  the  same  number  of  vertices  and  edges 
must  lie  on  each  face. 

All  the  edges  of  a  regular  polyhedron  are  equal. 

The  faces  of  a  convex  regular  polyhedron  are  all  convex 
polygons,  and  the  polyhedral  angles  are  all  convex. 

538.  At  each  vertex  there  must  meet  at  least  three  faces, 
and  in  each  face  there  must  lie  at  least  three  vertices. 

Moreover,  the  sum  of  the  face  angles  at  any  vertex  of  a 
convex  regular  polyhedron  is  less  than  four  right  angles,  or 
360  degrees.  (Art.  460.) 

From  these  two  properties  we  are  able  to  show  that  the 
greatest  possible  number  of  convex  regular  polyhedrons  is  five. 

The  faces  of  a  regular  convex  polyhedron  must  be  regular 
convex  polygons,  equilateral  triangles,  squares,  etc. 

1.  Since  the  interior  angle  of  an  equilateral  triangle  is  60 
degrees,  if  the  faces  of  a  regular  convex  polyhedron  are  equi- 
lateral triangles,  at  each  vertex  there  may  meet  three,  four,  or 
Jive  faces;  but  not  six. 

2.  Since  the  interior  angle  of  a  square  is  90  degrees,  if  the 
faces  of  a  regular  convex  polyhedron  are  squares,  at  each 
vertex  there  may  meet  three  faces,  but  not  four. 

3.  Since  the  interior  angle  of  a  regular  pentagon  is  108 
degrees,  if  the  faces  of  a  regular  convex  polyhedron  are  pen- 
tagons, at  each  vertex  there  may  meet  three  faces,  but  not  four. 

4.  The  interior  angle  of  a  regular  hexagon  is  120  degrees ; 
hence  it  is  impossible  for  three  or  more  regular  hexagons  to 


537-639] 


PBISMS  AND  PYEAMins 


343 


meet  at  a  vertex  so  as  to  form  a  convex  polyhedral  angle. 
Similarly  no  regular  polygon  of  more  than  six  sides  can  be 
a  face  of  a  regular  convex  polyhedron. 

Therefore  the  only  polygons  which  can  enter  into  the  con- 
struction of  regular  convex  polyhedrons  are  equilateral  triangles, 
squares,  and  regular  pentagons  and  the  ways  in  which  these  can 
be  combined  are : 

(1)  triangular  faces  meeting  three  at  a  vertex. 

(2)  triangular  faces  meeting  four  at  a  vertex. 

(3)  triangular  faces  meeting  Jive  at  a  vertex. 

(4)  square  faces  meeting  three  at  a  vertex. 

(5)  pentagonal  faces  meeting  three  at  a  vertex. 

What  is  here  shown  is  that  no  other  regular  convex  poly- 
hedrons than  the  five  answering  the  above  conditions  can 
exist;  that  these  five  varieties  do  exist  can  most  easily  be 
demonstrated  by  actually  making  models  of  them. 


r^ — V 


Tetrahedron 


Octahedron       Dodecahedron 


Icosahedron 


539. .  The  five  regular  polyhedrons  are : 

1.  The  regular  tetrahedron,  having  four  triangular  faces  meet- 
ing three  in  a  vertex. 

It  has  four  vertices  and  six  edges. 

2.  The  regular  octahedron,  having  eight  triangular  faces  meet- 
ing four  in  a  vertex. 

It  has  six  vertices  and  twelve  edges. 

3.  The  regular  icosahedron,  having  twenty  triangular  faces 
meeting  five  in  a  vertex. 

It  has  twelve  vertices  and  thirty  edges. 

4.  The  regular  hexahedron  or  cube,  having  six  square  faces 
meeting  three  in  a  vertex. 

It  has  eight  vertices  and  twelve  edges. 


344 


ELEMENTARY  GEOMETRY 


[Chap.  VII 


5.  The  regular  dodecahedron,  having  twelve  pentagonal  faces 
meeting  three  in  a  vertex. 

It  has  twenty  vertices  and  thirty  edges. 

540.  Models  of  the  five  regular  convex  polyhedrons  can 
easily  be  constructed  by  cutting  pieces  of  cardboard  in  the 
shape  of  the  diagrams  below,  and  folding  them  along  the 
dotted  lines  till  the  edges  come  together.  The  edges  should 
then  be  glued  or  pasted  over  with  strips  of  cloth  or  paper.  To 
make  the  folding  easy  it  would  be  well  to  cut  the  cardboard 
halfway  through  along  the  dotted  lines. 


Tetrahedron 


Octahedron 


Dodecahedron 


looaahedron 


539-541]  PRISMS  AND  PYRAMIDS  345 

Section  V 

POLYHEDRONS  IN  GENERAL 
Proposition  XX 

541.  In  any  polyhedron  the  numher  of  edges  increased 
hy  two  is  equal  to  the  number  of  faces  together  with  the 
number  of  vertices.    [Euler's  Theorem.] 

Let  E  be  the  number  of  edges  in  any  given  polyhedron,  F 
the  number  of  faces,  and  V  the  number  of  vertices. 

It  is  required  to  prove  that  E  -^2  =  F+  F. 

Proof.  Imagine  the  polyhedron  to  be  built  up  by  taking  one 
face  and  to  it  attaching  another,  and  another,  and  so  on,  till 
the  whole  figure  is  completed. 

Let  /  be  the  number  of  faces  of  the  incomplete  figure  at 
any  stage,  e  the  number  of  edges,  and  v  the  number  of  vertices. 

In  any  face  taken  alone,  the  number  of  vertices  equals  the 
number  of  edges. 

So  that  when/=  1,  e  =  u 

Now  to  the  first  face  add  a  second.  One  edge  of  the  second 
is  made  to  coincide  with  an  edge  of  the  first,  while  two  vertices 
of  the  second  coincide  with  two  vertices  of  the  first. 

That  is,  in  the  new  face  by  itself  the  number  of  edges  equals 
the  number  of  vertices,  but  when  this  face  is  attached  to  the 
one  already  taken,  two  vertices  are  lost,  while  only  one  edge  is 
lost. 

Therefore,  when  /=  2,      e  =  v  +  1. 

To  these  attach  a  third  face,  and  again  the  numtJer  of  vertices 
lost  will  exceed  the  number  of  edges  lost  by  one^ 

Therefore,  when  /  =  3,      e  =  v  +  2. 

Continue  this  process  till  all  the  faces  are  added  but  the 
last  one. 


346  ELEMENTARY  GEOMETRY  [Chap.  VII 

Then,  when  f=F—l,  e  =  v-{-(F—2),  since  the  number  of 
vertices  lost  for  each  of  the  F—2  faces  added  has  exceeded 
the  number  of  edges  lost  by  one. 

When  the  last  face  is  added,  all  of  its  edges  and  vertices  are 
made  to  coincide  with  edges  and  vertices  already  counted.  So 
that  adding  the  last  face  merely  completes  the  figure  without 
changing  the  relation  between  the  numbers  of  edges  and 
vertices. 

Therefore,  for  the  complete  figure, 

E=  r+(i^-2), 

or  E  +  2  =  V-\-F. 

Proposition  XXI 

542.  The  sum  of  the  face  angles  of  any  polyhedron, 
together  with  eight  right  angles,  is  equal  to  four  times 
as  many  right  angles  as  the  polyhedron  has  vertices. 

Let  F  denote  the  number  of  faces,  E  the  number  of  edges,  V 
the  number  of  vertices,  and  S  the  sum  of  the  face  angles  of 
any  polyhedron. 

It  is  required  to  prove  that  S  -{-S  right  angles  =  4  F  right 
angles. 

Proof.  In  any  face,  the  sum  of  the  face  angles  +  4  right 
angles  =  twice  as  many  right  angles  as  the  face  has  edges. 

(Art.  116.) 

Therefore,  adding  for  all  the  faces,  remembering  that  each 
edge  is  counted  twice, 

S  -\-^F  right  angles  =  2^x2  right  angles, 

or  S  =  4:(E  —  F)  right  angles 

=  4  (  F-  2)  right  angles,  (Prop.  XX.) 

i.e.  S  +  S  right  angles  =  4  Fright  angles. 


541-542]  PBISMS  AND  PYRAMIDS  347 


MISCELLANEOUS   EXERCISES 

1.  There  can  be  no  polyhedron  with  less  than  four  faces,  or  less  than 
six  edges. 

2.  No  polyhedron  can  be  constructed  having  seven  edges. 

3.  The  sum  of  the  areas  of  any  three  faces  of  a  tetrahedron  is  greater 
than  the  area  of  the  fourth  face. 

4.  If  a  tetrahedron  is  cut  by  a  plane  parallel  to  a  pair  of  opposite  edges, 
the  section  will  be  a  parallelogram. 

5.  The  straight  lines  which  join  the  mid-points  of  opposite  edges  of  a 
tetrahedron  are  concurrent. 

6.  If  a  perpendicular  is  drawn  from  a  vertex  of  a  regular  tetrahedron 
to  its  opposite  face,  show  that  the  foot  of  the  perpendicular  will  intersect 
each  median  of  that  face,  dividing  it  in  the  ratio  2:1. 

7.  Show  that  the  perpendicular  from  the  vertex  of  a  regular  tetrahedron 
upon  the  opposite  face  is  three  times  that  drawn  from  its  foot  to  any  of 
the  other  faces. 

8.  Prove  that  the  lateral  area  of  any  pyramid  is  greater  than  the  area 
of  the  base. 

Suggestion.  Project  the  vertex  on  the  base  and  join  its  projection  to 
the  vertices  of  the  base. 

9.  If  planes  are  passed  through  three  concurrent  edges  of  a  tetrahedron 
and  the  mid-points  of  their  opposite  edges,  they  will  intersect  in  one  straight 
line. 

10.  The  volume  of  a  triangular  prism  is  equal  to  one-half  the  product  of 
the  area  of  one  lateral  face  and  the  perpendicular  distance  of  it  from  the 
opposite-  edge. 

11.  The  plane  which  bisects  a  dihedral  angle  of  a  tetrahedron  cuts  the 
opposite  edge  into  segments  proportional  to  the  areas  of "  the  adjacent 
faces. 

12.  The  altitude  of  a  regular  tetrahedron  is  equal  to  the  sum  of  the 
four  perpendiculars  drawn  from  any  point  within  it  to  the  four  faces. 

13.  The  area  of  any  lateral  face  of  a  prism  is  less  than  the  sum  of  the 
areas  of  the  other  lateral  faces. 

14.  Show  that  the  six  planes  bisecting  the  dihedral  angles  of  any  tetra- 
hedron meet  in  one  point. 

•     Suggestion.     Compare  Ex.  1,  p.  303. 


348  ELEMENTARY  GEOMETRY  [Chap.  VII 

15.  The  six  planes  each  passing  through  one  edge  of  a  tetrahedron  and 
the  mid-point  of  the  opposite  edge,  meet  in  one  point. 

16.  Given  the  number  and  kind  of  faces  in  each  of  the  regular  poly- 
hedrons, and  the  number  of  faces  meeting  at  a  vertex,  compute  the  num- 
ber of  vertices  and  edges. 

For  example,  in  the  regular  dodecahedron  there  are  twelve  pentagonal 
faces  meeting  three  in  a  vertex. 

12  faces,  5  vertices  in  each  face  gives  60  face  angles. 

fin 
3  face  angles  at  each  vertex,  gives  —  =  20  vertices. 

o 

Also  12  faces,  5  edges  in  each  face  gives  60  edges, 

but  this  counts  each  edge  twice,  since  each  edge  lies  in  two  faces. 

60 
Therefore  the  actual  number  of  edges  is  —  =  30  edges. 

Ji 

17.  Verify  Euler's  formula,  E  -\- 2  =  V-\-  F,  for  the  case  of  each  of  the 
regular  convex  polyhedrons. 

18.  Verify  Euler's  formula  for  the  case  of  a  hexagonal  pyramid  and  a 
hexagonal  prism. 

19.  The  total  number  of  plane  angles  in  the  faces  of  any  polyhedron 
is  even,  and  equal  to  twice  the  number  of  edges. 

20.  The  homologous  edges  of  two  similar  tetrahedrons  are  in  the  ratio 
5 :  6.     Find  the  ratio  of  their  total  areas  and  their  volumes. 

21.  One  edge  of  a  tetrahedron  is  6  inches.  Find  the  length  of  the 
homologous  edge  of  a  similar  tetrahedron  having  (1)  twice  the  volume, 
(2)  twice  the  total  area. 

22.  Two  parallelepipeds  are  similar  if  the  three  faces  meeting  at  any 
vertex  are  similar  and  similarly  placed. 

23.  If  four  straight  lines,  of  which  no  three  lie  in  the  same  plane,  in- 
tersect in  one  point,  and  are  cut  by  two  parallel  planes  on  opposite  sides 
of  their  common  point,  show  that  the  eight  points  of  intersection  are  the 
vertices  of  a  parallelepiped. 

24.  The  base  of  a  pyramid  is  12  square  feet  and  its  altitude  is  6  feet. 
Find  the  area  of  a  section  parallel  to  the  base  and  2  feet  from  it. 

25.  The  bases  of  a  frustum  of  a  pyramid  are  8  square  feet  and  4J 
square  feet,  respectively,  and  its  altitude  is  5  feet.    Find  its  volume. 

26.  A  cistern  is  8  feet  square  at  the  top,  6  feet  square  at  the  bottom, 
and  8  feet  deep.  How  many  gallons  of  water  will  it  hold,  assuming  7| 
gallons  to  the  cubic  foot  ? 


Summary]  PRISMS  AND  PYRAMIDS  349 

SUMMARY  OF   CHAPTER  VII 
1,   Definitions. 

(1)  Plane  Section  of  a  Surface  — the  points  common  to  a  surface 

and  an  intersecting  plane.     §  468. 

(2)  Closed  Surface  —  a  surface  such  that  every  plane  section  of  it 

consists  of  one  or  more  closed  lines.     §  469. 

(3)  Polyhedron  —  a  surface  made  up  of  intersecting  planes.     §  470. 

(4)  Convex  Polyhedron  —  one  which  lies  wholly  on  one  side  or  the 

other  of  each  of  its  faces.     §  470. 

(5)  Tetrahedron,  Pentahedron,  Hexahedron,  Octahedron,  Dodecahe- 

dron, Icosahedron.     See  §  470. 

(6)  Prism  —  a  polyhedron  of  which  two  faces  are  convex  polygons 

lying  in  parallel  planes  and  identically  equal,  while  the  remain- 
ing faces  are  parallelograms.     §  471. 

(7)  RigJit  Prism  —  one  whose  lateral  edges  are  perpendicular  to  its 

bases.     §  472. 

(8)  Right  Section  of  a  Prism  —  a  section  made  by  a  plane  perpen- 

dicular to  the  lateral  edges.     §  473. 

(9)  Lateral  Area  of  a  Prism  —  the  sum  of  the  areas  of  the  lateral 

faces.    §  474. 

(10)  Altitude  of  a  Prism  — the  perpendicular  distance  between  its 

bases.     §  475. 

(11)  Truncated  Prism  —  a  polyhedron  having  parallel  lateral  edges 

like  a  prism,  but  bases  which  are  neither  parallel  nor  equal. 
§  483. 

(12)  Volume  of  a  Polyhedron  —  the  space  enclosed  by  the  polyhedron, 

or  the  measure  of  that  space.     §§  485,  502. 

(13)  Parallelepiped  —  a  prism  whose  bases  are  parallelograms.     §  489. 

(14)  Right  Parallelepiped  —  one  in  which  a  set  of  lateral  edges  is 

perpendicular  to  the  bases.     §  490. 

(15)  Rectangular  Parallelepiped  —  a,  right  parallelepiped  whose  bases 

are  rectangles.     §  491. 

(16)  Cube  —  a  parallelepiped  whose  faces  are  all  squares.     §  492. 

(17)  Pyramid  —  a  polyhedron  one  of  whose  faces  is  a  polygon,  and 

the  remaining  faces  are  triangles  having  a  common  vertex. 
§509. 

(18)  Altitude  of  a  Pyramid  — the  perpendicular  distance  from  the 

vertex  to  the  base.     §  509. 

(19)  Regular  Pyramid— one  whose  base  is  a  regular  polygon  and 

whose  vertex  lies  on  the  perpendicular  to  the  base  drawn 
from  its  centre.     §  510. 


350  ELEMENTARY  GEOMETRY  [Chap.  VIi 

(20)  Slant  Height  of  a  Regular  Pyramid  —  the  altitude  of  any  of  its 

lateral  faces.     §  510. 

(21)  Truncated  Pyramid  —  the  figure  determined  by  the  base  and  any 

section  of  a  pyramid  and  the  portions  of  the  lateral  faces  inter- 
cepted between  them.     §  511. 

(22)  Frustum  of  a  Pyramid  —  a  truncated  pyramid  whose  bases  are 

parallel.     §  511. 

(23)  Similar  Polyhedrons  —  two  polyhedrons  which  have  the  same 

number  of  faces  similar,  each  to  each,  and  similarly  placed, 
and  have  their  corresponding  polyhedral  angles  equal.     §  530. 

(24)  Regular  Polyhedron  —  one  whose  faces  are  equal  regular  poly- 

gons, and  whose  polyhedral  angles  are  all  equal.     §  537. 

2.  Axioms. 

(1)  If  two  figures  are  identically  equal,  their  volumes  are  equal 

(Axiom  13).      §  486. 

(2)  If  equal  volumes  be  added  to  equal  volumes,  or  to  the  same 

volume,  their  sums  are  equal  (Axiom  14).     §  487. 

3.  Theorems  on  the  Equality  of  Prisms. 

(1)  Two  prisms,   or  truncated  prisms,  are  identically  equal  if  the 

three  faces  forming  a  trihedral  angle  in  one  are  identically 
equal  to  the  three  faces  forming  a  trihedral  angle  in  the  other, 
and  are  similarly  placed.     §§  481,  484. 

(2)  Two  right  prisms  are  identically  equal  if  their  bases  are  identically 

equal  and  they  have  equal  altitudes.     §  482. 

4.  Theorems  on  the  Sections  of  Prisms  and  Pyramids. 

(1)  The  sections  of  a  prism  made  by  parallel  planes  are  polygons 

which  are  identically  equal.     §  476. 

(2)  If  a  pyramid  is  cut  by  a  plane  parallel  to  its  base,  the  edges  and 

the  altitude  are  divided  in  the  same  ratio,  the  section  made  by 
the  plane  is  similar  to  the  base,  and  the  area  of  the  section  is 
to  the  area  of  the  base  as  the  squares  of  their  distances  from 
the  vertex.     §  513. 

(3)  If  two  pyramids  have  equal  altitudes  and  bases  of  equal  areas, 

the  areas  of  sections  made  by  planes  equidistant  from  their 
vertices  are  equal.    §  516. 

6.    Theorems  on  the  Lateral  Areas  of  Prisms  and  Pyramids. 

(1)  The  lateral  area  of  a  prism  is  equal  to  the  product  of  a  lateral 
edge  and  the  perimeter  of  a  right  section  of  the  prism.     §  479. 


Summary]  PBISMS  AND  PYRAMIDS  351 

(2)  The  lateral  area  of  a  right  prism  is  equal  to  the  product  of  its 

altitude  aud  the  perimeter  of  its  base.     §  480. 

(3)  The  lateral  area  of  a  regular  pyramid  is  equal  to  one-half  the 

product  of  the  perimeter  of  the  base  and  the  slant  height. 
§512. 

6.  Theorems  on  the  Volume  of  a  Prism. 

•  (1)  An  oblique  prism  is  equal  in  volume  to  a  right  prism  whose  base 
is  a  right  section  of  the  oblique  prism  and  whose  altitude  is 
equal  to  a  lateral  edge  of  the  oblique  prism.     §  488. 

(2)  The  volumes  of  two  rectangular  parallelepipeds  having  identically 

equal  bases  are  in  the  same  ratio  as  their  altitudes.     §  496. 

(3)  The  volumes  of  two  rectangular  parallelepipeds  having  equal 

altitudes  are  in  the  same  ratio  as  the  areas  of  their  bases. 
§499. 

(4)  The  volume  of  any  rectangular  parallelepiped  is  the  product  of 

its  altitude  and  the  area  of  its  base,     §  504. 

(5)  The  volume  of  any  parallelepiped  is  equal  to  the  product  of  its 

altitude  and  the  area  of  its  base.     §  505. 

(6)  The  volume  of  a  prism  is  equal  to  the  product  of  its  altitude  and 

the  area  of  its  base.     §§506,  507. 

(7)  The  volume  of  a  truncated  triangular  prism  is  equal  to  the  sum 

of  the  volumes  of  three  pyramids  whose  common  base  is  one 
base  of  the  prism  and  whose  vertices  are  the  three  vertices  of 
the  other  base.     §  527. 

(8)  The  volume  of  a  truncated  right  triangular  prism  is  equal  to  the 

product  of  the  area  of  that  face  to  which  the  edges  are  perpen- 
dicular and  one-third  the  sum  of  the  lateral  edges.     §  528. 

(9)  The  volume  of  any  truncated  triangular  prism  is  equal  to  the 

product  of  the  area  of  a  right  section  and  one -third  the  sum  of 
the  lateral  edges.     §  529. 

7.  Theorems  on  the  Volume  of  a  Pyramid. 

(1)  If  two  triangular  pyramids  have  equal  altitudes  and  equal  bases, 

they  have  equal  volumes.     §  517. 

(2)  The  volume  of  a  triangular  pyramid  is  one-third  the  volume  of  a 

triangular  prism  of  the  same  base  and  altitude.     §  521. 

(3)  The  volume  of  a  triangular  pyramid  is  equal  to  one-third  the 

product  of  its  altitude  and  the  area  of  its  base.     §  522. 

(4)  The  volume  of  any  pyramid  is  equal  to  one-third  the  product  of 

its  altitude  and  the  area  of  its  base.     §  523. 


352  ELEMENTARY  GEOMETRY  [Chap.  VII 

(5)  The  volumes  of  any  two  pyramids  are  in  the  same  ratio  as  the 

products  of  their  altitudes  and  the  areas  of  their  bases.     §  524. 

(6)  The  volumes  of  two  pyramids  having  equal  altitudes  are  in  the 

same  ratio  as  the  areas  of  their  bases,  or  having  equal  bases 
are  in  the  same  ratio  as  their  altitudes.     §  525. 

(7)  The  volumes  of  two  tetrahedrons  having  a  trihedral  angle  of  the 

one  equal  to  a  trihedral  angle  of  the  other  are  in  the  same  ratio 
as  the  products  of  the  edges  which  meet  in  the  vertices  of  these 
angles.     §  526. 

8.  Theorems  on  Similar  Polyhedrons. 

(1)  Any  two  edges  or  diagonals  of  a  polyhedron  are  in  the  same  ratio 

as  the  homologous  edges  or  diagonals  of  a  similar  polyhedron. 
§531. 

(2)  The  surfaces  of  two  similar  polyhedrons  {i.e.  the  sums  of  the 

areas  of  all  the  faces)  are  in  the  same  ratio  as  the  squares  of 
any  two  homologous  edges  or  homologous  diagonals.     §  532. 

(3)  Two  tetrahedrons  are  similar  if  the  three  faces  meeting  in  a 

vertex  of  the  one  are  similar  and  similarly  situated  to  the 
three  faces  meeting  in  a  vertex  of  the  other.     §  533. 

(4)  The  volumes  of  two  similar  tetrahedrons  are  in  the  same  ratio  as 

the  cubes  of  their  homologous  edges.     §  535. 

(5)  The  volumes  of  any  two  similar  polyhedrons  are  in  the  same  ratio 

as  the  cubes  of  two  homologous  edges.     §.53G. 

9.  Miscellaneous  Theorems. 

(1)  The  opposite  lateral  faces  of  a  parallelepiped  are  parallel  and 

Identically  equal.     §  493. 

(2)  Any  two  opposite  faces  of  a  parallelepiped  may  be  taken  as  the 

bases.     §  494. 

(3)  The  plane  passed  through  two  diagonally  opposite  edges  of  a 

parallelepiped  divides  it  into  two  triangular  prisms  which  are 
equal  in  volume.    §  495. 

(4)  In  any  polyhedron  the  number  of  edges  increased  by  two  is  equal 

to  the  number  of  faces  together  with  the  number  of  vertices 
(Euler's  Theorem).    §  541. 

(5)  The  sum  of  the  face  angles  of  any  polyhedron  together  with  eight 

right  angles  is  equal  to  four  times  as  many  right  angles  as  the 
polyhedron  has  vertices.     §  542. 


CHAPTER   VIII 
CYLINDERS  AND  CONES 

Section  I 

CYLINDERS 

543.  Definitions.  If  a  straight  line  moves  so  as  always  to 
remain  parallel  to  a  fixed  straight  line,  while  some  point  of  it 
traverses  a  fixed  curve  not  in  a  plane  with  the  fixed  line,  it  is 
said  to  describe  a  cylindrical  surface. 


Cylindrical  sitrface 

Each  position  of  the  moving  line  is  parallel  to  its  former 
position,  hence  the  line  is  said  to  move  parallel  to  itself. 

The  moving  line  is  called  the  generator  of  the  cylindrical 
surface,  and  the  guiding  curve,  the  director. 
2  A  353 


354 


ELEMENTARY  GEOMETRY 


[Chap.  VIII 


The  generator  in  any  position  is  an  element  of  the  surface. 
A  closed  cylindrical  surface  is  one  whose  section,  made  by  any 
plane  cutting  all  of  its  elements,  is  a  closed  curve. 


544.  A  cylinder  is  a  figure  consisting  of 
two  parallel  plane  surfaces  and  a  closed 
cylindrical  surface  intercepted  between 
them. 

The  two  plane  surfaces  are  called  the 
bases,  and  the  cylindrical  surface,  the 
lateral  surface  of  the  cylinder.  Through 
every  point  of  the  lateral  surface  there 
passes  an  element  of  the  surface. 

A  circular  cylinder  is  one  whose  bases 
are  circular. 

A  cylinder  is  called  a  right  cylinder  when  the  elements  of  its 
lateral  surface  are  perpendicular  to  the  bases ;  otherwise  it  is 
an  oblique  cylinder. 


Oblique  cylinder 


545.   A  right  circular  cylinder  may  be  generated  by  revolv- 
ing a  rectangle  about  one  of  its  sides.      The  opposite  side 
will    then   generate   the   cylindrical   surface, 
and    the    adjacent    sides    will    generate   the 
bases. 

For  this  reason  a  right  circular  cylinder  is 
sometimes  called  a  cylinder  of  revolution. 

The  side  of  the  rectangle  about  which  the 
rotation  takes  place  is  called  the  axis  of  the 
cylinder. 


Right  cylinder 


546.   A  right  section  of  a  cylinder  is   the 

section   made   by   a  plane   perpendicular  to 
the  elements  of  the  lateral  surface. 

The  altitude  of  a  cylinder  is  the  perpendicular  distance  between 
its  bases. 


543-550] 


CYLINDERS  AND  CONES 


355 


547.  Theorem  1.  If  a  plane  contains  one  element  of  the 
lateral  surface  of  a  cylinder  and  meets  the  lateral  surface  at  any 
other  point,  it  contains  also  a  second  element,  and  the  section  of 
the  cylinder  by  the  plane  is  a  parallelogram. 

If  the  plane  L  contains  the  element  AB  of  the  lateral  surface  of  the 
cylinder  PQ,  and  meets  the  lateral  surface  again 
at  D,  it  must  contain  also  the  element  through 
Z>,  since  that  element  is  parallel  to  AB^  and 
must  lie  in  a  plane  with  AB. 

The  plane  L  therefore  cuts  the  lateral  surface 
of  the  cylinder  along  two  parallel  lines  AB  and 
CD,  and  the  bases  along  two  straight  lines  AD 
and  BC  which  are  also  parallel  (Art.  419). 

Hence  the  section  of  the  cylinder  made  by  the 
plane  is  a  parallelogram. 

548.  Corollary.  Tlie  section  of  a 
right  cylinder  made  by  a  plane  containing 
an  element  of  the  lateral  surface  is  a 
rectangle. 

549.  Definition.  If  a  plane  contains  one  element  of  the 
lateral  surface  of  a  cylinder,  and  only  one,  it  is  said  to  be 
tangent  to  the  cylinder. 

The  element  which  the  tangent  plane  contains  is  called  the 
element  of  contact. 

A  straight  line  is  tangent  to  a  cylinder  when  it  meets  an 
element  of  the  lateral  surface,  and  lies  in  the  tangent  plane 
containing  that  element. 

550.  Theorem  II.  Tlie  plane  determined  by  a  tangent  to  any 
circular  section  of  a  cylinder  and  the  element  of  the  lateral  sur- 
face passing  through  its  point  of  contact  is  tangent  to  the  cylinder. 

For  if  it  is  not  tangent  to  the  cylinder  it  must  contain  a  second 
element  of  the  lateral  surface,  and  so  contain  a  chord  of  the 
section  instead  of  the  tangent  to  the  section. 

Conversely.  If  a  plane  is  tangent  to  a  circular  cylinder  it 
intersects  the  plane  of  the  base  in  a  tangent  to  the  base. 


356 


ELEMENTARY  GEOMETRY 


[Chap.  VIII 


551.  Definition.  If  sections  of  a  cylindrical  surface  are 
made  by  two  planes,  those  are  corresponding  points  of  the 
sections  which  lie  upon  the  same  element  of  the  surface. 

Proposition  I 

552.  Parallel  sections  of  a  cylindrical  surface  are 
identically  equal. 


Let  QR  be  any  cylindrical  surface  of  which  PABC  and 
P'A'B'C  are  parallel  sections. 

It  is  required  to  prove  that  PABC  and  P'A'B'C  are  identi- 
cally equal. 

Proof.  Suppose  A,  A' ;  B,  B' ;  C,  C  are  any  three  pairs  of 
corresponding  points  in  the  two  sections. 

Then,  AA'  and  BB'  are  not  only  parallel  but  also  equal 
(Art.  420),  and  hence  ABB' A'  is  a  parallelogram.      (Art.  127.) 


Therefore 
Similarly, 


AB  =  A'B'. 


AC=A'C',2iSidL 


(Art.  123.) 


BC=B'0'. 


551-556]  CYLINDERS  AND   CONES  357 

Therefore  As  ^50  and  A'B'C  are  identically  equal,  and 
can  be  superposed.  (Art.  53.) 

Suppose  now  that  the  section  P'A'B'C  is  placed  upon  the 
section  PABC  so  that  the  points  A',  B',  C  coincide  with  their 
corresponding  points  A,  B,  C,  and  let  A,  A'  and  C,  C  remain 
fixed  while  B  traverses  the  section  PABC.  Since  B  will 
always  coincide  with  its  corresponding  point  B',  every  point 
of  the  one  section  will  coincide  with  its  corresponding  point 
in  the  other. 

Therefore  the  two  sections  are  identically  equal. 

553.  Corollary  I.  The  bases  of  a  cylinder  are  identically 
equal. 

554.  Corollary  II.  A7iy  two  right  sectioiis  of  a  cylinder  are 
identically  equal. 

555.  Corollary  III.  Any  section  of  a  cylinder  parallel  to 
the  base  is  identically  equal  to  the  base. 

556.  Corollary  IV.  All  sections  of  a  circular  cylinder 
parallel  to  its  bases  are  equal  circles,  and  the  straight  line  joining 
the  centres  of  the  bases  passes  through  the  centres  of  all  the  parallel 
sections. 

EXERCISES 

1.  If  through  any  point  of  the  lateral  surface  of  a  cylinder  a  straight 
line  is  drawn  parallel  to  any  element  of  the  surface,  this  straight  line  is 
also  an  element  of  the  surface. 

2.  Show  that  all  the  elements  of  the  lateral  surface  of  a  cylinder  are  of 
equal  lengths. 

3.  The  line  of  intersection  of  two  planes  tangent  to  a  cylinder  is 
parallel  to  an  element  of  the  lateral  surface,  and  to  the  plane  through 
their  two  elements  of  contact. 

4.  Find  the  locus  of  points  (1)  at  a  given  distance  from  a  given  straight 
line  ;  (2)  at  given  distances  from  each  of  two  given  parallel  straight  lines. 


358 


ELEMENTARY  GEOMETRY 


[Chap.  VIII 


Definitions 

557.   A  prism  is  inscribed  in   a  cylinder 

when  its  bases  and  the  bases  of  the  cylin- 
der lie  in  the  same  planes,  and  its  lateral 
edges  are  elements  of  the  lateral  surface  of 
the  cylinder. 

The  cylinder  is  at  the  same  time  circum- 
scribed about  the  prism. 

The  section  of  an  inscribed  prism  made 
by  any  plane  is  inscribed  in  the  section  of 
the  cylinder  made  by  the  same  plane. 


Priam  inscribed  in 
a  cylinder 


558.   A  prism  is  circumscribed  about 

a  cylinder  when  its  bases  and  the 
bases  of  the  cylinder  lie  in  the  same 
planes,  and  its  lateral  faces  are  tan- 
gent to  the  cylinder. 

The  cylinder  is  at  the  same  time 
inscribed  in  the  prism. 

The  section  of  a  circumscribed 
prism  made  by  any  plane  is  circum- 
scribed about  the  section  of  the  cylin- 
der made  by  the  same  plane. 


559.  If  a  regular  prism  is  inscribed, 
or  circumscribed,  to  a  circular  cylin- 
der, and  the  number  of  its  lateral  faces 
is  indefinitely  increased  in  some  regu- 
lar way,  the  lateral  surface  of  the  prism  approaches  the  lateral 
surface  of  the  cylinder  as  its  limit,  the  volume  of  the  prism 
approaches  the  volume  of  the  cylinder  as  its  limit,  and  the 
perimeter  and  area  of  a  right  section  of  the  prism  approach 
the  perimeter  and  area  of  a  right  section  of  the  cylinder  as 
their  limits. 


Prism  circmnscribed  about  a 
cylinder 


657-561] 


CYLINDERS  AND  CONES 


359 


Proposition  II 

560.  The  lateral  area  of  a  circular  cylinder  is  equal 
to  the  product  of  the  perimeter  of  a  right  section  of  the 
cylinder  and  the  length  of  an  element  of  its  lateral 
surface. 


Let  PQ  be  a  circular  cylinder  of  which  S  is  the  perimeter  of 
a  right  section,  and  AB  an  element  of  the  lateral  surface. 

It  is  required  to  prove  that  the  lateral  area  of  PQ  is  equal  to 
S  times  AB. 

Proof.  In  the  cylinder  PQ  inscribe  a  regular  prism  AH, 
one  of  whose  elements  is  AB.  Let  KM  be  a  right  section  of 
this  prism.     The  lateral  area  of  AH=  perimeter  of  KM  x  AB. 

(Art.  479.) 

If  the  number  of  lateral  faces  of  AH  is  indefinitely  increased, 
the  lateral  area  of  AH  approaches  the  lateral  area  of  PQ  as 
its  limit,  and  the  perimeter  of  KM  approaches  S  as  its  limit. 

Therefore  the  lateral  area  of  PQ  =  S  x  AB.  (Art.  230.) 

561.  Corollary.  The  lateral  area  of  a  right  circular  cylin- 
der is  equal  to  the  product  of  the  altitude  and  the  circumference  of 
the  base. 

If  the  lateral  area  is  represented  by  A,  the  radius  of  the  base 
by  r,  and  the  altitude  by  h, 

A  =  2  Tvh. 


360  ELEMENTARY  GEOMETRY  [Chap.  VIII 


Proposition  III 

562.  The  volume  of  a  circular  cylinder  is  equal  to  the  product 
of  its  altitude  and  the  area  of  its  base. 

The  proof  of  this  theorem  is  similar  to  that  of  Proposition  II, 
and  is  left  to  the  pupil.  Reference  should  be  made  to 
Article  507. 

If  the  volume  is  represented  by  F,  the  altitude  by  h,  and  the 
radius  of  the  base  by  r, 

V  =  irr^h. 

563.  Corollary.  Tfie  volumes  of  all  circular  cylinders, 
whether  right  or  oblique,  having  equal  bases  and  equal  altitudes 
are  equal. 

EXERCISES 
Note.     Use  3^  as  the  approximate  value  of  rr. 

1.  Find  the  volume,  the  lateral  area,  and  the  total  area  of  a  right  cir- 
cular cylinder  the  diameter  of  whose  base  is  14  inches,  and  whose  altitude 
is  11  inches. 

2.  The  lateral  area  of  a  right  circular  cylinder  is  528  square  feet,  and 
its  volume  is  1584  cubic  feet.  Find  the  diameter  and  circumference  of 
its  base,  and  its  altitude. 

3.  Two  right  circular  cylinders  are  of  equal  height  while  the  circum- 
ference of  one  is  double  the  circumference  of  the  other.  What  is  the 
ratio  of  their  lateral  areas  ?    Also  of  their  volumes  ? 

4.  A  hollow  cylindrical  iron  tube  has  an  outer  diameter  of  12  inches 
and  an  inner  diameter  of  9  inches.  How  many  cubic  inches  of  iron  are 
there  in  a  piece  of  the  tube  2  feet  long  ? 

5.  When  a  tap  is  opened  the  water  in  a  pipe  of  one  inch  inner  diam- 
eter flows  at  the  rate  of  two  miles  an  hour.  How  many  gallons  of  water 
would  flow  from  the  tap  in  20  minutes  ? 

Take  1\  gallons  to  a  cubic  foot. 

6.  Show  that  the  volumes  of  two  similar  right  circular  cylinders  are  in 
the  same  ratio  as  the  cubes  of  their  altitudes,  or  as  the  cubes  of  the  radii 
of  their  bases. 

7.  What  is  the  altitude  of  a  right  circular  cylinder  if  its  lateral  area 
equals  the  sum  of  the  areas  of  its  bases  ? 


662-565]  CYLINDERS  AND   CONES  361 

Section  II 

THE  CONE 

564.  Definition.  If  a  straight  line  moves  so  as  always 
to  pass  through  a  fixed  point,  while  some  point  of  the  line 
traverses  a  fixed  curve  not  in  a  plane  with  the 

fixed  point,  it  will  describe  a  conical  surface.  m^_   ^^ 

As  in  the  cylindrical  surface,  the  generating       W~     "W^ 
line  in  any  position  is  an  element  of  the  surface.  \     W 

The  directing  curve  will  always  be  considered  \X 

a  closed  curve.  /\ 

The  fixed  point  through  which  the  generating  /    \ 

line  always  passes  is  called   the  vertex  of  the       J       Ik 
surface.  m"        1^ 

If  the  generating  line  is  indefinite  in  length, 
the  surface  is  divided  into  two  parts  at  the  ver-       ^"*^"  mrface 
tex,  called  the  two  sheets,  or  the  two  nappes  of  the  conical  surface. 

When  we  speak  of  a  pla.ne  section  of  a  conical  surface,  we 
shall  always  have  in  mind  the  section  made  by  a  plane  which 
cuts  all  the  elements  on  the  same  side  of  the  vertex,  that  is,  a 
section  of  one  of  the  sheets. 

565.  Definition.  A  cone  is  a  figure  consisting  of  a  plane 
surface  and  a  conical   surface    intercepted 

between  the  plane  surface  and  the  vertex.  a 

The  plane  surface  is  called  the  base,  and  /  \ 

the  conical  surface  the  lateral  surface.     The  £'      \ 

vertex  of  the  conical   surface  is  the  vertex  M          m, 

of  the  cone.  ^^_ ]«. 

The  straight  line  joining  any  point  of  the     ^P^  ^^m 

lateral  surface  to  the  vertex  is  an  element  of     ^^«1.^__^ \^ 

the  lateral  surface,  since  it  must  coincide  circular  cone 

with  the  generating  line  in  one  position. 

A  circular  cone  is  one  whose  base  is  circular.  The  straight 
line  joining  the  vertex  to  the  centre  of  the  base  is  called  the 
axis  of  the  cone. 


362 


ELEMENTARY  GEOMETRY 


[Chap.  VIII 


When  the  vertex  lies  on  the  perpendicular  to  the  base  drawn 
from  its  centre,  the  cone  is  called  a  right  circular  cone. 

566.  The   altitude  of  a  cone  is  the  perpendicular  distance 
from  the  vertex  to  the  plane  of  the  base. 

567.  The  slant  height  of  a  right  circular  cone  is  the  length  of 
an  element  of  the  lateral  surface. 


568.  Theorem  I.    Any  plane  which  contains  one  element  of  the 
lateral  surface  of  a  cone  and  meets  the  sur- 
face at  any  other  point,  contains  also  a  second 
elemeyit,  and  the  section  of  the  cone  by  the 
plane  is  a  triangle. 

If  the  plane  L  contains  the  element  SB  of  the 
cone  S-ABC,  and  meets  the  surface  again  at  the 
point  D  it  must  contain  the  element  SD,  since  it 
contains  two  points  of  this  element. 

The  section  of  the  cone  by  the  plane  L  there- 
fore consists  of  the  line-segments  SB,  SD,  and 
BD,  and  is  thus  a  triangle. 

569.  Corollary.  The  section  of  a  right 
circular  cone  by  a  plane  through  the  vertex  is 
an  isosceles  triangle. 

570.  Definition.  If  a  plane  contains  one  element  of  the 
lateral  surface  of  a  cone  and  only  one,  it  is  said  to  be  tangent 
to  the  cone,  and  the  element  which  it  contains  is  called  the 
element  of  contact. 

571.  Theorem  II.  The  plane  determined  by  a  tangent  to  the 
base  of  a  circular  cone  and  the  element  of  the  lateral  surface 
passing  through  its  point  of  contact  is  tangent  to  the  cone. 

Conversely.     If  a  plane  is  tangent  to  a  circular  cone,  its 
intersection  with  the  plane  of  the  base  is  tangent  to  the  base. 
See  Article  550. 


565-573]  CYLINDERS  AND   CONES  363 

Proposition  IV 
572.  t^ny  section  of  a  circular  cone  made  hy  a  plane 
parallel  to  the  base  is  a  circle,  and  its  centre  lies  upon 
the  straight  line  joining  the  vertex  to  the  centre  of  the 
hose. 


77" "' "  -  *"  X 7 


Let  A'B'Q'  be  the  section  of  the  circular  cone  S-ABQ  made 
by  a  plane  L  parallel  to  the  base,  and  let  SO  he  the  straight 
line  joining  S  to  the  centre  of  the  base. 

It  is  required  to  prove  that  A'B'Q'  is  a  circle  whose  centre 
lies  on  SO. 

Proof.  Let  A'  and  B'  be  any  two  points  of  the  section,  and 
let  the  planes  determined  by  the  line  SO  and  A',  SO  and  B' 
intersect  the  lateral  surface  of  the  cone  in  the  straight  lines 
SA'A,  SB'B,  the  base  in  the  lines  AO,  BO,  and  the  plane  L  in 
the  lines  A'O',  B'O',  parallel  to  J.0  and  BO,  respectively. 

Therefore  ^^^^  =  9^.  (Art.  242.) 

OA       OS       OB  ^  ^ 

But  OA  =  OB.     Therefore  0'A'=0'B'. 
Since  A'  and  B'  are  any  points  whatever  in  the  section  made 
by  L,  all  points  of  the  section  are  equidistant  from  0'. 
Therefore  the  section  is  a  circle  with  its  centre  at  0'. 

573.  Corollary.  The  radii  of  two  sections  of  a  circular 
cone,  parallel  to  the  base,  are  in  the  same  ratio  as  their  distances 
from  the  vertex. 


364 


ELEMENTARY  GEOMETRY 


[Chap.  VIII 


574.  Definitions.  A  pyramid  is  inscribed  in  a  cone  when 
their  bases  lie  in  the  same  plane,  and  the  lateral  edges  of  the 
pyramid  are  elements  of  the  lateral  surface  of  the  cone. 


Pyramid  inscribed 
in  a  cone 


Pyramid  circumHcrihed 
about  a  cone 


The  vertex  of  the  inscribed  pyramid  coincides  with  the 
vertex  of  the  cone,  and  the  base  of  the  pyramid  is  inscribed  in 
the  base  of  the  cone.  The  section  of  the  inscribed  pyramid 
made  by  any  plane  is  inscribed  in  the  section  of  the  cone  made 
by  the  same  plane. 

575.  A  pyramid  is  circumscribed  about  a  cone  when  their  bases 
lie  in  the  same  plane,  and  the  lateral  faces  of  the  pyramid  are 
tangent  to  the  cone. 

The  vertex  of  the  circumscribed  pyramid  coincides  with  the 
vertex  of  the  cone,  and  the  base  of  the  pyramid  is  circum- 
scribed about  the  base  of  the  cone.  The  section  of  the  circum- 
scribed pyramid  made  by  any  plane  is  circumscribed  about 
the  section  of  the  cone  made  by  the  same  plane. 


576.  If  a  regular  pyramid  is  inscribed,  or  circumscribed,  to 
a  circular  cone  and  the  number  of  its  lateral  faces  is  indefinitely 
increased  in  some  regular  way,  the  lateral  surface  of  the  pyr- 
amid approaches  the  lateral  surface  of  the  cone  as  its  limit, 
and  the  volume  of  the  pyramid  approaches  the  volume  of  the 
cone  as  its  limit. 


574-578] 


CYLINDERS  AND   CONES 


365 


Proposition  V 

577.  The  lateral  area  of  a  right  circular  cone  is  equal 
to  one-half  the  product  of  the  circumference  of  the  base 
and  the  slant  height. 

S 


Let  S-ACE  be  a  right  circular  cone,  of  which  I  is  the  slant 
height,  and  M  the  circumference  of  the  base. 

It  is  required  to  prove  that  the  lateral  area  of  S-ACE  is 
equal  to  \  Ml. 

Proof.  Inscribe  in  the  cone  S-ACE  a  regular  pyramid  P, 
one  of  whose  elements  is  SA  =  I,  and  the  perimeter  of  whose 
base  is  H. 

Then  the  lateral  area  oi  P=\Hl  (Art.  512.) 

If  the  number  of  lateral  faces  of  P  is  indefinitely  increased, 
the  lateral  area  of  P  approaches  the  lateral  area  of  S-ACE  as 
its  limit,  and  the  perimeter  H  approaches  the  circumference  M 
as  its  limit. 

Therefore  the  lateral  area  of  S-ACE  =  \  Ml         (Art.  230.) 

If  the  radius  of  the  base  is  r,  and  the  lateral  area  is  represented 

by  A 

A  =  irrl. 


578.  Definition.  The  frustum  of  a  cone  is  that  portion 
of  a  cone  intercepted  between  the  base  and  a  plane  parallel 
to  the  base  intersecting  the  lateral  surface. 


366  ELEMENTARY   GEOMETRY  [Chap.  VIII 


Proposition  VI 

579.   The  volume  of  a  circular  cone  is  equal  to  one- 
third  the  product  of  its  altitude  and  the  area  of  its 

base. 

The  proof  of  this  theorem  is  similar  to  that  of  Proposition  V, 

and  is  left  to  the  pupil. 
If  the  volume  is  represented  by  F,  the  altitude  by  ^,  and  the 

radius  of  the  base  by  r,  * 

V=  \TrrVi. 


MISCELLANEOUS    EXERCISES 

1.  Find  the  lateral  area  and  the  volume  of  a  right  circular  cone,  the 
area  of  whose  base  is  154  square  inches  and  whose  altitude  is  11  mches. 

2.  The  slant  height  of  a  right  circular  cone  is  4  metres.  How  far  from 
the  vertex  must  a  section  parallel  to  the  base  be  taken  so  as  to  divide  the 
lateral  area  into  two  equal  parts  ? 

3.  A  right  triangle  whose  sides  are  3,  4,  5  feet,  respectively,  is  rotated 
about  the  shortest  side.  Find  the  area  of  the  surface  described  by  the 
hypotenuse. 

4.  Show  that  the  lateral  areas  of  similar  right  circular  cylinders  are 
in  the  same  ratio  as  the  squares  of  their  altitudes,  or  as  the  squares  of  the 
radii  of  their  bases. 

Definition.  Similar  right  circular  cylinders  are  generated  by  the 
rotation  of  similar  rectangles  about  homologous  sides; 

6.  Show  that  the  lateral  areas  of  two  similar  cones  of  revolution  are 
in  the  same  ratio  as  the  squares  of  their  slant  heights,  or  as  the  squares 
of  their  altitudes,  or  as  the  squares  of  the  radii  of  their  bases. 

Definition.  Similar  cones  of  revolution,  or  right  circular  cones,  are 
generated  by  similar  right  triangles  rotating  about  homologous  sides. 

6.  Show  that  the  volumes  of  two  similar  right  circular  cones  are  in  ■ 
the  same  ratio  as  the  cubes  of  their  altitudes,  or  as  the  cubes  of  the  radii 
of  their  bases. 

7,  The  volumes  of  two  similar  cones  of  revolution  are  in  the  ratio  of 
512  .  729.     What  is  the  ratio  of  their  lateral  areas  ?. 


579]  CYLINDERS  AND  CONES  367 

8.  Show  that  the  lateral  area  of  the  frustum  of  a  right  circular  cone  is 
equal  to  the  sum  of  the  circumferences  of  its  bases  multiplied  by  one- half 
the  slant  height. 

9.  Show  that  the  lateral  area  of  the  frustum  of  a  right  circular  cone  is 
equal  to  the  circumference  of  a  section  midway  between  the  bases  multi- 
plied by  the  slant  height. 

10.  Show  that  the  volume  of  the  frustum  of  a  circular  cone,  the  areas 
of  whose  bases  are  B  and  b  and  whose  altitude  is  h,  is  given  by  the 
formula 

V=lh(B+b-hVBb). 
See  Ex.  2,  p.  338. 

11.  A  right  circular  cylinder  of  height  2  ft.  and  the  radius  of  whose 
base  is  6  in.  rolls  on  a  plane  making  one  complete  revolution.  What  is 
the  shape  of  the  surface  covered  by  it  ?    Find  its  area. 

12.  A  right  circular  cone  whose  altitude  is  12  in.  and  the  radius  of  whose 
base  is  5  in.  rolls  on  a  plane,  without  slipping,  making  one  complete  revo- 
lution.    What  is  the  shape  of  the  surface  covered  ?    Find  its  area. 

13.  A  regular  hexagonal  prism  is  inscribed  in  a  right  circular  cylinder. 
Compare  their  lateral  areas  and  their  volumes. 

14.  The  base  of  a  right  circular  cylinder  has  a  radius  of  7  cm.  and  an 
altitude  of  7  cm.    Find  its  total  surface  area  and  its  volume. 

15.  A  rectangle  whose  adjacent  sides  are  5  ft.  and  7  ft.,  respectively, 
revolves  in  succession  about  these  sides;  show  that  the  volumes  of  the 
cylinders  generated  are  in  the  ratio  7  :  5. 

16.  The  total  surface  of  a  right  circular  cone  is  462  sq.  cm.,  and  the 
slant  height  is  twice  the  radius  of  the  base  ;  find  the  volume. 

17.  In  a  right  circular  cylinder  of  height  3  ft.  and  8  in.  radius,  a  square 
prism  is  inscribed.     Find  its  volume. 

18.  A  square  whose  side  is  50  cm.  revolves  about  one  of  its  diagonals  ; 
find  the  area  and  volume  of  the  figure  so  generated. 

19.  Show  that  in  a  right  circular  cone  all  the  elements  of  the  lateral 
surface  are  equal  in  length. 

20.  Show  that  the  total  area  T  of  a  right  circular  cylinder,  including 
the  areas  of  the  two  bases,  is  given  by  the  formula  : 

T=2  irrh  +  2  Trr^=:  2irr(h-\-  r). 

21.  A  circular  cistern  is  22  ft.  in  circumference  at  the  top,  16  ft.  at  the 
bottom,  and  8  ft.  deep.  How  many  gallons  of  water  will  it  hold,  assum- 
ing 7 1  gallons  to  the  cubic  foot  ? 


368  ELEMENTARY  GEOMETRY  [Chap.  VIII 

SUMMARY   OF   CHAPTER   VIII 

1.  Definitions. 

(1)  Cylindrical  Surface  —  the  surface  described  by  a  straight  line 

moving  parallel  to  a  fixed  straight  line,  while  some  point  of  it 
traverses  a  fixed  curve  not  in  a  plane  W\t\\  the  fixed  line.     §  543. 

(2)  Conical  Surface  —  the  surface  described  by  a  straight  fine  which 

moves  so  as  always  to  pass  through  a  fixed  point,  while  some 
point  of  it  traverses  a  fixed  curve  not  in  a  plane  with  the  fixed 
point.     §  564. 

(3)  Closed  Cylindrical  Surface  —  one  for  which  the  directing  curve 

is  closed.     §  543. 

(4)  Vertex  of  a  Conical  Surface  —  the  point  through  which  all  the 

elements  of  the  surface  pass.     §  564. 

(5)  Cylinder  —  a  figure  consisting  of  two  parallel  plane  surfaces  and 

a  closed  cylindrical  surface  intercepted  between  them.     §  544. 

(6)  Circular  Cylinder  —  one  whose  bases  are  circular.     §  544. 

(7)  Bight  Cylinder  —  one  in  which  the  elements  of  the  lateral  surface 

are  perpendicular  to  the  bases.     §  544. 

(8)  Cone  —  a  figure  consisting  of  a  plane  and  the  conical  surface 

intercepted  between  it  and  the  vertex.     §  565. 

(9)  Circidar  Cone  —  one  whose  base  is  circular.     §  565. 

(10)  Bight  Circular  Cone  —  one  whose  vertex  lies  on  the  perpen- 

dicular to  the  base  drawn  from  its  centre.     §  565. 

(11)  Tangent  to  a  Cylinder  or  a  Cone  —  a  plane  containing  one  element 

of  the  lateral  surface,  and  only  one  ;  also,  a  line  meeting  one 
element  of  the  surface,  and  only  one.     §  549.     §  570. 

(12)  Frustum  of  a  Cone  —  that  portion  of  a  cone  intercepted  between 

its  base  and  a  plane  parallel  to  the  base,  intersecting  the  lateral 
surface.     §  578. 

2.  Theorems  on  the  Properties  of  Cylinders. 

(1)  A  plane  which  contains  one  element  of  the  lateral  surface  of  a 

cylinder  in  general  contains  also  another  element,  and  the 
section  is  a  parallelogram.     §  547. 

(2)  The  section  of  a  right  cylinder  made  by  a  plane  containing  an 

element  of  the  lateral  surface  is  a  rectangle.     §  548. 

(3)  The  plane  determined  by  a  tangent  to  any  circular  section  of  a 

cylinder  and  the  element  of  the  lateral  surface  passing  through 
its  point  of  contact  is  tangent  to  the  cylinder  ;  and  conversely, 
if  a  plane  is  tangent  to  a  circular  cylinder  it  intersects  the 
plane  of  the  base  in  a  tangent  to  the  base.     §  550. 


Summary]  CYLINDERS  AND   CONES  369 

(4)  Parallel  sections  of  a  cylindrical  surface  are  identically  equal. 

§552. 

(5)  The  bases  of  a  cylinder  are  identically  equal.    §  553. 

(6)  Any  two  right  sections  of  a  cylinder  are  identically  equal.    §  554. 

(7)  Any  section  of  a  cylinder  parallel  to  the  base  is  identically  equal 

to  the  base.     §  555. 

(8)  All  the  sections  of  a  circular  cylinder  parallel  to  its  bases  are 

equal  circles,  and  the  straight  line  joining  the  centres  of  the 
bases  passes  through  the  centres  of  all  the  parallel  sections. 
§566. 

3.  Theorems  on  the  Properties  of  Cones. 

(1)  Any  plane  containing  one  element  of  the  lateral  surface  of  a  cone 

contains  in  general  a  second  element  and  the  section  of  the  cone 
by  the  plane  is  a  triangle.     §  568. 

(2)  The  plane  determined  by  a  tangent  to  the  base  of  a  circular  cone 

and  the  element  of  the  lateral  surface  passing  through  its  point 
of  contact  is  tangent  to  the  cone ;  and  conversely,  if  a  plane  is 
tangent  to  a  circular  cone,  its  intersection  with  the  plane  of 
the  base  is  tangent  to  the  base.     §  571. 

(3)  Any  section  of  a  circular  cone  made  by  a  plane  parallel  to  the 

base  is  a  circle,  and  its  centre  lies  upon  the  straight  line  joining 
the  vertex  to  the  centre  of  the  base.     §  572. 

4.  Theorems  on  the  Lateral  Area  and  the  Volume  of  a  Cylinder. 

(1)  The  lateral  area  of  a  circular  cylinder  is  equal  to  the  product  of 

the  perimeter  of  a  right  section  of  the  cylinder  and  the  length 
of  an  element  of  its  lateral  surface.     §  560. 

(2)  The  lateral  area  of  a  right  circular  cylinder  is  equal  to  the  product 

of  the  circumference  of  the  base  and  the  altitude.  A  =  2  irrh. 
§561. 

(3)  The  volume  of  a  circular  cylinder  is  equal  to  the  product  of  its 

altitude  and  the  area  of  its  base,     V=  irr^h.     §  562. 

(4)  The  volumes  of  all  circular  cylinders,  whether  right  or  oblique, 

having  equal  bases  and  equal  altitudes  are  equal.     §  563. 

5.  Theorems  on  the  Lateral  Area  and  the  Volume  of  a  Cone. 

(1)  The  lateral  area  of  a  right  circular  cone  is  equal  to  one-half  the 

product  of  the  circumference  of  the  base  and  the  slant  height. 
A  =  Trrh.     §577. 

(2)  The  volume  of  a  circular  cone  is  equal  to  one-third  the  product 

of  its  altitude  and  the  area  of  its  base.      V=\  irr^h.     §  579. 
2b 


CHAPTER   IX 
THE  SPHERE 

Section  I 

PLANE   SECTIONS   AND   TANGENT  PLANES 

580.  Definitions.  A  sphere  is  a  closed  surface  such  that 
all  points  of  it  are  equidistant  from  a  fixed  point  within  it. 

The  fixed  point  is  called  the  centre  of  the  sphere. 

If  a  circle  is  rotated  about  its  diameter,  the  surface  described 
by  it  is  a  sphere. 

Any  straight  line  drawn  from  the  centre  to  a  point  on  the 
sphere  is  called  a  radius  of  the  sphere. 

A  straight  line  drawn  through  the  centre  and  terminated  both 
ways  by  the  sphere  is  called  a  diameter  of  the  sphere. 

The  length  of  a  diameter  is  twice  the  length  of  a  radius. 

It  follows  from  the  definition  of  a  sphere  that  all  radii  of  the 
same  sphere  are  equal,  and  hence  that  all  diameters  of  the  same 
sphere  are  equal. 

581.  Theorem.  Equal  spheres  have  equal  radii  and  equal 
diameters. 

Conversely.  Spheres  which  have  equal  radii  or  equal  diame- 
ters are  equal. 

582.  Spheres  which  have  the  same  centre  but  do  not  coincide 
are  called  concentric  spheres. 

Spheres  having  the  same  centre  and  one  point  in  common 
must  coincide  throughout. 

370 


680-685]  THE  SPHERE  371 


Proposition  I 

583.   The  section  of  a  sphere  made  by  a  plane  is  a 
circle. 


Let  0  be  tlie  centre  of  the  given  sphere  and  A^  B,  C  three 
points  on  the  section  made  by  any  plane  L. 

It  is  required  to  prove  that  ABC  is  a  circle. 

Proof.  From  0  draw  OP  perpendicular  to  the  plane  L,  draw 
the  radii  OA  and  OB,  and  join  PA  and  PB. 

As  OP  A  and  OPB  are  identically  equal.  (Art.  78.) 

Hence  PA  =  PB. 

But  A  and  B  are  any  two  points  common  to  the  plane  and 
sphere. 

Therefore  all  points  common  to  the  plane  and  sphere  are 
equidistant  from  P. 

Therefore  the  section  of  the  sphere  made  by  the  plane  Z  is  a 
circle. 

584.  Corollary  I.  Sections  of  a  sphere  made  by  planes 
equidistant  from  the  centre  are  equal  circles,  and  of  two  sections 
made  by  planes  unequally  distant  from  the  ceritre,  that  is  the 
greater  circle  which  is  made  by  the  nearer  plane. 

585.  Definition.  Any  circle  lying  on  a  sphere  is  called 
a  circle  of  the  sphere. 

A  circle  whose  plane  passes  through  the  centre  of  the  sphere 
is  called  a  great  circle  of  the  sphere;  one  whose  plane  does 
not  pass  through  the  centre  is  called  a  small  circle  of  the  sphere. 


372  ELEMENTARY  GEOMETRY  [Chap.  IX 

A  quadrant  is  one  quarter  of  a  great  circle  of  a  sphere,  and 
subtends  a  right  angle  at  the  centre  of  the  sphere. 

586.  Corollary  II.  TJie  centre  of  any  circle  of  a  sphere  is 
the  foot  of  the  pei'pendicular  drawn  from  the  centre  of  the  sphere 
to  the  plane  of  the  circle. 

587.  Corollary  III.  The  centres  of  all  great  circles  coin- 
cide with  the  centre  of  the  sphere^  and  all  great  circles  on  the  same 
sphere  are  equal. 

588.  Corollary  IV.  Every  great  circle  divides  the  sphere 
into  two  equal  parts. 

589.  Corollary  V.  Any  two  great  circles  on  the  same  sphere 
bisect  each  other. 

Their  planes  intersect  in  a  diameter  of  each  circle. 

590.  Corollary  VI.  An  arc  of  a  great  circle  can  he  drawn 
through  any  two  given  points  on  a  sphere,  and  if  the  two  given 
points  are  not  the  extremities  of  a  diameter,  only  one  such  arc  less 
than  a  semicircle  can  he  drawn. 

The  two  given  points  and  the  centre  of  the  sphere  determine 
the  plane  of  the  great  circle. 

591.  Corollary  VII.  TJirough  three  given  points  on  a  sphere 
one  and  only  one  circle  can  he  drawn. 

592.  Definition.  The  diameter  of  a  sphere  perpendicular 
to  the  plane  of  any  circle  of  the  sphere  is  called  the  axis  of  that 
circle,  and  the  extremities  of  the  diameter  are  called  the  poles 
of  the  circle. 

Are  the  poles  of  a  circle  of  a  sphere  equally  or  unequally  dis- 
tant from  the  plane  of  the  circle  ? 

593.  By  the  distance  between  two  points  on  a  sphere  is 
meant  the  length  of  the  arc,  not  greater  than  a  semicircle,  of 
the  great  circle  passing  through  them. 

A  distinction  is  made  here  between  the  distance  measured  on 
a  sphere  and  measured  along  a  straight  line. 


585-596] 


THE  SPHERE 


373 


Proposition  II 

594.  All  points  of  any  circle  of  a  sphere  are  equidis- 
tant from  either  of  its  poles. 


Let  A,  B,  C  be  any  three  points  of  a  given  circle  of  the 
sphere  whose  centre  is  0,  and  let  P  and  P'  be  the  poles  of 
that  circle. 

It  is  required  to  prove  that  A,  B^  C  are  equidistant  from  P, 
and  also  equidistant  from  P'. 

Proof.  If  planes  are  passed  through  the  line  PP^  and  the 
points  A,  B,  C,  the  arcs  determined  by  them,  PA,  PB,  PC, 
P'A,  P'B,  P'C,  are  arcs  of  great  circles.  Hence  the  distance 
from  P  to  J.  is  the  length  of  the  arc  PA,  etc. 

Since  PP'  is  perpendicular  to  the  plane  of  the  circle  ABC  at 
its  centre,  the  chords  PA,  PB,  PC  are  all  equal.     Prove. 

Therefore  the  arcs  PA,  PB,  PO'are  all  equal.         (Art.  163.) 

Similarly,  the  arcs  PA,  P'B,  P'C  are  all  equal. 


595.  Definition.  The  distance  on  a  sphere  of  any  point 
of  a  circle  from  its  nearer  pole  is  called  the  polar  distance  of 
the  circle. 

596.  Corollary  I.  TTie  polar  distance  of  any  point  of  a 
great  circle  is  one-fourth  the  circumference  of  a  great  circle,  or 
more  briefly,  is  a  quadrant. 


374 


ELEMENTABY  GEOMETRY 


IJhap.  IX  J 


597.  Corollary  II.  If  a  point  on  a  sphere  is  a  quadrant's 
distance  from  each  of  two  given  points  on  the  sphere,  it  is  the  pole 
of  the  great  circle  passing  through  the  two 
points. 

Let  P  be  a  quadrant's  distance  from  A  and 
B ;  throuj?h  PA,  PB,  and  AB  pass  great  circles. 
Then  Z  POA  is  a  right  angle.  Also  Z  POB  is  a 
right  angle.  Therefore  PO  is  perpendicular  to 
the  plane  AOB.  (Art.  397.)  Hence  P  is  the 
pole  of  the  great  circle  through  A  and  B. 


Corollary  III.  If  one  point  on  a  sphere  is  a  quad- 
rant's distance  from  another,  it  is  the  pole  of  some  great  circle 
passing  through  the  other. 


599.  The  study  of  the  geometry  of  a  sphere,  or  Spherical  Geometry, 
is  greatly  simplified  by  the  use  of  a  slated  globe  upon  which  chalk  marks 
can  be  made  and  erased.  If  the  length  of  the  radius  of  such  a  spherical 
surface  is  known,  Corollary  II  enables  us  to  draw  on  it  the  arc  of  a 
great  circle  passing  through  any  two  given  points. 

To  do  this  first  draw  in  any  plane  a  circle 
whose  radius  r  is  equal  to  the  radius  of  the 
given  spherical  surface,  and  make  an  opening 
in  the  compasses  equal  to  EF.,  the  chord  of 
a  quadrant. 

Then,  if  A  and  B  are  the  given  points  on 
the  spherical  surface,  arcs  described  on  the 
sphere,  with  the  stationary  arm  of  the  com- 
passes at  A  and  B  in  turn,  will  intersect  at  P, 
the  pole  of  the  great  circle  through  A  and  B. 
The  arc  described  with  the  stationary  arm  at  P,  and  with  the  same  open- 
ing, will  then  be  the  arc  of  a  great  circle  through  A  and  B. 

By  the  same  method  we  can  find  the  pole  of  a  given  arc  of  a  great 
circle,  or  draw  the  great  circle  of  which  a  given  point  is  the  pole. 

A  geometrical  method  of  finding  the  length  of  the  diameter  of  a  given 
material  spherical  solid  will  be  found  in  Proposition  XIX,  page  393. 

For  all  work  on  a  sphere,  a  pair  of  compasses  with  curved  arms  will  be 
found  most  convenient. 


597-601]  THE  SPHERE  375 

600.  Definitions.  A  plane  which  meets  a  sphere  at  one 
and  only  one  point  is  tangent  to  the  sphere.  The  point  which 
the  sphere  and  plane  have  in  common  is  called  the  point  of 
contact. 

A  straight  line  is  tangent  to  a  sphere  when  it  meets  the  sphere 
at  one  and  only  one  point,  no  matter  how  far  it  is  produced. 

A  straight  line  tangent  to  a  sphere  is  tangent  to  every  sec- 
tion of  the  sphere  made  by  planes  through  the  line. 

Proposition  III 

601.  A  plane  which  is  perpendicular  to  a  radius  of 
a  sphere  at  its  extremity  is  tangent  to  the  sphere. 


Let  the  plane  L  be  perpendicular  to  the  radius  OA  of  the 
sphere  whose  centre  is  0,  at  its  extremity  A. 

It  is  required  to  prove  that  the  plane  L  is  tangent  to  the 
sphere. 

Proof.  First,  the  plane  has  the  point  A  in  common  with 
the  sphere. 

Next,  every  other  point  of  L  lies  outside  of  the  sphere. 
Why? 

Therefore  the  plane  L  is  tangent  to  the  sphere. 

Conversely.  A  plane  tvhich  is  tangent  to  a  sphere  is  perpen- 
dicular to  the  radius  drawn  to  its  point  of  contact. 

For  if  not,  the  perpendicular  to  it  drawn  from  the  centre  is  less 
than  a  radius,  and  the  plane  must  pass  inside  the  sphere. 


376  ELEMENTARY   GEOMETRY  [Chap.  IX 

602.  Corollary  I.  All  lines  tangent  to  a  sphere  at  one 
point  lie  in  a  plane  tangent  to  the  sphere  at  that  point. 

For  they  are  all  perpendicular  to  the  radius  through   the 
point.     Why  ? 

603.  Corollary  II.  TTie  plane  of  two  straight  lines  tangent 
to  a  sphere  at  the  same  point  is  also  tangent  to  the  sphere  at  that 
point. 

Proposition  IV 

604.  The  intersection  of  two  spheres  is  a  circle,  the 
plane  of  which  is  perpendicular  to  the  straight  line 
joining  the  centres  of  the  spheres,  and  the  centre  of 
which  is  on  that  line. 


Let  0  and  0'  be  the  centres  of  two  spheres  which  intersect. 

It  is  required  to  prove  that  their  intersection  is  a  circle 
whose  plane  is  perpendicular  to  00'  and  whose  centre  lies 
on  00\ 

Proof.     Let  A  be  any  point  common  to  the  two  spheres. 

Through  A  and  00'  pass  a  plane.  This  will  cut  the  spheres 
in  two  circles,  which  intersect  at  A  and  another  point  B. 

The  chord  AB  is  perpendicular  to  00',  and  is  bisected  by 
it  at  C.  .   (Art.  168.) 

If  the  plane  of  OAO'  is  revolved  about  00',  the  two  circles 
will  describe  the  given  spheres,  and  the  point  A  will  trace 
their  line  of  intersection. 

Since  AC  is  perpendicular  to  00'  and  remains  of  constant 
length,  A  will  describe  a  circle  whose  plane  is  perpendicular 
to  00'  and  whose  centre  is  C. 


602-606]  THE  SPHERE  377 


Proposition  V 

605.   Through  any  four  points  not  lying  in  the  same 
plane,  one  and  only  one  sphere  can  be  passed. 


Let  A,  B,  C,  D  be  four  points  not  lying  in  the  same  plane. 

It  is  required  to  prove  that  one  and  only  one  sphere  can  be 
passed  through  A^  B,  C,  and  D. 

Proof.  Pass  planes  through  the  points  A,  B,  C  and  A,  D,  C, 
and  find  the  centres  M  and  N  of  the  circles  circumscribing 
As  ABC  and  ADC,  respectively.  (Art.  151.) 

Let  K  be  the  mid-point  of  the  line-segment  AC,  and  join 
MK  and  NK 

These  lines  will  be  perpendicular  to  ^C  at  K        (Art.  166.) 

The  plane  MKN  is  perpendicular  to  both  planes  ABC  and 
ADC,  being  perpendicular  to  their  intersection  AC.     (Art.  438.) 

The  perpendiculars  to  the  planes  ABC  and  ADC  erected  at 
M  and  N  will  therefore  lie  in  the  plane  MKN,  and  must  inter- 
sect at  some  point  0.  (Arts.  440,  98.) 

Then  0  is  the  centre  of  a  sphere  which  passes  through  the 
four  points  A,  B,  C,  D. 

For,  the  straight  line  MO  is  the  locus  of  points  equidistant 
from  A,  B,  and  C. 

The  straight  line  NO  is  the  locus  of  points  equidistant  from 
A,  D,  and  C. 


378  ELEMENTARY  GEOMETRY  [Chap.  IX 

Hence  the  point  0  is  equidistant  from  A,  B,  C,  and  D,  and 
a  sphere  whose  centre  is  0  and  radius  equal  to  OA  will  pass 
through  the  four  given  points. 

Also,  since  MO  and  NO  have  no  other  common  point  than  O, 
no  other  sphere  can  be  passed  through  the  four  given  points. 

606.  Definition.  A  polyhedron  is  said  to  be  inscribed  in  a 
sphere  (or  the  sphere  to  be  circumscribed  about  the  polyhedron) 
when  the  vertices  of  the  polyhedron  lie  on  the  sphere.  A 
polyhedron  is  said  to  be  circumscribed  about  a  sphere  (or  the 
sphere  to  be  inscribed  in  the  polyhedron)  when  the  faces  of 
the  polyhedron  are  tangent  to  the  sphere. 

607.  Corollary  I.  One  and  only  one  sphere  can  be  circuTn- 
scribed  about  a  tetrahedron. 

608.  Corollary  II.  TTie  perpendiculars  to  the  four  faces  of 
a  tetrahedron,  erected  at  the  centres  of  the  circumscribed  circleSy 
all  pass  through  one  point. 

609.  Corollary  III.  The  six  planes  perpendicular  to  the 
edges  of  a  tetrahedron  at  their  mid-points  have  one  point  in 
common. 

EXERCISES 

1.  If  from  a  point  outside  of  a  sphere  straight  lines  are  drawn  tangent 
to  the  sphere,  the  line-segments  lying  between  the  point  and  the  points  of 
contact  are  all  equal. 

2.  Show  that  in  the  preceding  example  the  points  of  contact  all  lie  on 
a  small  circle  of  the  sphere  whose  axis  is  the  straight  line  joining  the 
given  point  to  the  centre  of  the  sphere. 

3.  Construct  a  sphere  which  will  pass  through  the  eight  vertices  of  a 
cube. 

4.  Show  that  the  sphere  which  is  passed  through  two  pairs  of  diagonally 
opposite  vertices  of  a  rectangular  parallelepiped  passes  also  through  the 
other  vertices. 


605-611]  THE  SPHERE  379 


Pboposition  VI 

610.   One  and   only  one  sphere  can  be  inscribed  in 
any  given  tetrahedron. 

D 


Let  ABCD  be  any  given  tetrahedron. 

It  is  required  to  prove  that  one  and  only  one  sphere  can  be 
inscribed  in  ABCD. 

The  proof  is  left  to  the  pupil. 

Suggestion.     Pass  planes  through  the  three  edges  of  any  one  side,  as 
ABC,  bisecting  the  dihedral  angles. 

611.   Corollary.     The  planes  ivhich  bisect  the  six  dihedral 
angles  of  any  tetrahedron  have  one  point  in  common. 

EXERCISES 

1.  What  is  the  locus  of  points  at  a  distance  a  from  a  point  A,  and  at  a 
distance  h  from  a  point  B  ? 

2.  What  is  the  locus  of   points  in  space  whose   distances  from  two 
fixed  points  are  in  a  given  ratio  ?     (See  Prop.  XX,  Chap.  III.) 

3.  What  is  the  locus  of  the  centres  of  the  spheres  to  which  three  given 
mtersecting  planes  are  tangent  ? 

4.  If  two  spheres  are  tangent  to  the  same  plane  at  the  same  point,  the 
straight  line  joining  their  centres  passes  through  that  point. 

5.  How  many  spheres  can  be  made  to  touch  four  planes  not  all  passing 
through  the  same  point  ? 


380  ELEMENTARY  GEOMETRY  [Chap.  IX 

Section  II 

SPHERICAL  ANGLES,  TRLA.NGLES,   AND  POLYGONS 

612.  Definition.  The  angle  formed  by  any  two  intersect- 
ing circles  is  defined  to  be  the  angle  formed  by  their  tangents 
at  their  common  point. 

The  same  definition  holds  for  the  angle  formed  by  any  two 
intersecting  curves. 

The  angle  formed  by  two  intersecting  arcs  of  great  circles  of 
a  sphere  is  called  a  spherical  angle. 

When  the  tangents  to  the  two  arcs  are  at  right  angles  the 
spherical  angle  is  called  a  right  spherical  angle. 

The  point  at  which  the  two  arcs  intersect  is  called  the 
vertex  of  the  spherical  angle. 

613.  The  planes  of  two  great  circles  of  a  sphere  intersect 
in  a  diameter.      Hence  the  tangents  of 

two  intersecting  arcs  of  great  circles  at 
their  common  point  are  perpendicular 
to  the  edge  of  the  dihedral  angle  formed 
by  their  planes,  and  the  angle  between 
the  tangents  is  the  plane  angle  of  the 
dihedral  angle  formed  by  the  planes  of 
the  circles.  (Art.  430.) 

Therefore 

Theorem  I.  The  angle  formed  by  two  intersectiyig  arcs  of 
great  circles  has  the  same  measure  as  the  dihedral  angle  formed 
by  the  planes  of  the  circles. 

614.  Suppose  PA  and  PB  are  two  given  arcs  of  great 
circles.  Draw  the  great  circle  of  which  P  is  the  pole,  and  let 
it  intersect  PA  at  Q,  and  PB  at  R.  Join  QO  and  RO,  0  being 
the  centre  of  the  sphere. 

Since  PQ  is  a  quadrant,  Z.POQ  is  a  right  angle.  Similarly 
Z  POR  is  a  right  angle.     Therefore  Z  QOR  is  the  plane  angle 


612-615]  THE  SPHERE  381 

of  the  dihedral  angle  formed  by  the  planes  of  FA  and  PB, 
and  is  equal  to  the  angle  between  the  tangents  at  P,  i.e.  to  the 
spherical  angle  APB. 

Now  Z  QOR :  four  right  angles  =  arc  QM :  a  great  circle. 

(Art.  267.) 
rpi        p  spherical  angle  APB      _       arc  QR 

four  right  spherical  angles      a  great  circle 

Since  the  second  and  fourth  terms  in  this  proportion  are 
constant  quantities,  the  above  relation  is  sometimes  briefly 
expressed  by  saying  that  "  the  spherical  angle  APB  Is  propor- 
tioned to  the  arc  Qi?,"  meaning  thereby  that  any  change  in  the 
one  causes  a  proportional  change  in  the  other. 

Therefore 

Theorem  II.  The  angle  formed  by  two  intersecting  arcs  of 
great  circles  is  proportional  to  the  arc  of  the  great  circle  of 
ivhich  the  vertex  of  the  angle  is  the  pole,  intercepted  between  the 
given  arcs  (produced  if  necessary). 

615.    Suppose  we  choose  any  unit  of  measure  for  angles,  as  for  ex- 
ample, a  degree,  and  take  as  unit  of  measure  for  arcs  that  arc  of  a  great 
circle  which  subtends  a  unit  angle  at  the  cen- 
tre of  the  sphere. 

Thus  if  ABC  is  a  great  circle  of  a  sphere, 
and  AOB  is  a  unit  angle,  the  arc  AB  is  taken 
as  the  unit  arc. 

This  unit  arc  is  also  frequently  called  a 
'  degree,'  but  care  must  be  taken  to  distinguish 
between   'angle    degree'    and   'arc    degree.' 

Arc  degrees  are  all  equal  in  the  same  circle, 
or  in  equal  circles,  but  are  unequal  in  unequal 
circles. 

Any  quadrant  of  a  circle  contains  90  arc  degrees,  a  semicircle  180  arc 
degrees,  and  a  circle  360  arc  degrees. 

In  the  diagram  of  Art.  613,  Z  QOR  will  have  the  same  measure  as  the 
arc  QR,  the  units  of  measure  being  chosen  as  above.  But  the  spherical 
angle  QPR  has  the  same  measure  as  Z  QOR.  Therefore  the  spherical 
angle  QPR  has  the  same  measure  as  the  arc  QR. 

Hence  Theorem  II  can  be  stated  as  follows  : 


382  ELEMENTARY  GEOMETRY  [Chap.  IX 

Theorem  II.  The  angle  formed  by  two  intersecting  arcs  of 
great  circles  has  the  same  measure  as  the  arc  of  the  great  circle 
of  which  the  vertex  of  the  angle  is  the  pole,  intercepted  between  the 
given  arcs  {produced  if  necessary). 

616.  If,  in  the  diagram  of  Article  613,  Q  is  the  pole  of  the 
great  circle  PSP',  as  it  is  the  pole  of  some  great  circle  through 
P  (Prop.  II,  Cor.  Ill),  then  the  spherical  angle  PQS  has  the 
same  measure  as  the  arc  PS. 

But  the  arc  PS  is  a  quadrant,  hence  /.PQS  is  a  right 
spherical  angle.  Now  PQ  is  any  arc  drawn  from  P  to  the  great 
circle  of  which  P  is  the  pole. 

Therefore 

Theorem  III.  All  arcs  of  great  circles  drawn  through  the  pole 
of  a  given  great  circle  are  perpendicular  to  the  given  great  circle. 

617.  Definition.  A  spherical  polygon  is 
a  closed  figure  on  a  sphere  consisting  of  arcs 
of  great  circles  which  intersect,  two  and  two, 
in  order. 

The  points  in  which  the  arcs  intersect  are 
the  vertices  of  the  polygon,  the  arcs  inter- 
cepted between  consecutive  vertices  are  the 
sides,  and  the  spherical  angles  formed  by  consecutive  sides  are 
the  angles  of  the  polygon. 

A  spherical  polygon  of  three  sides  is  called  a  spherical 
triangle. 

A  convex  spherical  polygon  lies  wholly  on  one  side  or  the 
other  of  each  of  its  sides  produced  ever  so  far. 

618.  Theorem  IV.  No  side  of  a  convex  spherical  polygon 
can  be  greater  than  a  semicircle. 

For,  if  the  side  AD  of  the  spherical  polygon  ABCD  in  the 
diagram  is  greater  than  a  semicircle,  AB  produced  must  inter- 
sect AD  between  A  and  Z),  since  great  circles  of  a  sphere  bisect 


615-621]  THE  SPHERE  383 

each  other.  In  that  case  the  polygon  would  lie  partly  on  one 
side  of  AB  and  partly  on  the  other,  and  hence  could  not  be 
convex. 

619.  Definition.  Two  spherical  polygons  are  identically 
equal  when  they  can  be  made  to  occupy  the  same  position,  i.e. 
when  they  can  be  superposed. 

620.  The  planes  of  the  great  circles,  of  which  the  sides  of  a 
spherical  polygon  are  arcs,  form  at  the  centre  of  the  sphere  a 
polyhedral  angle  whose  face  angles  have  the  same  numerical 
measures  as  the  sides  of  the  polygon,  and  whose  dihedral  angles 
have  the  same  measures  as  the  angles  of  the  polygon. 

.  If  the  spherical  polygon  is  convex,  the  corresponding  poly- 
hedral angle  is  convex,  and  conversely. 

Two  spherical  polygons  are  identically  equal  if  the  cor- 
responding polyhedral  angles  are  equal,  and  conversely. 

Many  properties  of  spherical  polygons  can  be  derived  directly 
from  known  properties  of  the  corresponding  polyhedral  angles. 

Proposition  VII 

621.  The  sum  of  any  two  sides  of  a  spherical  triangle 
is  greater  than  the  third  side. 


Suggestions  for  Proof.  The  sum  of  any  two  face  angles  of  a  tri- 
hedral angle  is  greater  than  the  third  face  angle.  (Art.  459.) 

How  do  the  measures  of  the  sides  of  the  spherical  triangle  compare 
with  the  measures  of  the  face  angles  of  the  corresponding  trihedral 
angle  ? 


384  ELEMENTARY  GEOMETRY  [Chap.  IX 


Proposition  VIII 

622.    The  sum  of  the  sides  of  a  convex  spherical  poly- 
gon is  less  than  a  great  circle. 

For  the  sum  of  the  face  angles  of  the  corresponding  polyhedral 
angle  is  less  than  four  right  angles.  (Art.  460.) 


Proposition  IX 

623.  //  two  spherical  triangles  lying  on  the  same 
sphere,  or  on  equal  spheres,  have  the  three  sides  of  one 
equal,  respectively,  to  the  three  sides  of  the  other,  their 
corresponding  angles  are  also  equal. 

Compare  Proposition  XXIV,  Chapter  VI. 

624.  It  should  be  carefully  noticed  that  while  two  mutually 
equilateral  spherical  triangles  are  also  mutually  equiangular, 
they  are  not  necessarily  superposable. 

Suppose  that  the  two  triangles  ABC  y<^  "y^ 

and  A'B'C  lie  on  the  same  sphere,  and 
are  such  that  the  straight  lines  AA', 
BE',  CC  are  diameters  of  the  sphere. 
The  two  trihedral  angles  formed  at  0 
have  their  face  angles  respectively  equal, 
and  likewise  their  dihedral  angles  equal ; 
but  they  are  not  identically  equal  since 
their  parts  are  arranged  in  opposite  orders  around  the  vertex. 

So  also  the  triangles  ABC  and  A'B'C  are  mutually  equi- 
lateral, and  likewise  mutually  equiangular,  but  their  parts  are 
arranged  in  opposite  orders. 

Such  triangles  cannot  be  superposed,  but  are  symmetrical. 

If  A  A'B'C  were  moved  on  the  surface  of  the  sphere  so  as 
to  make  the  vertices  A'  and  B'  coincide  with  A  and  B,  respec- 


622-627]  THE  SPHERE  385 

tively,  the  vertex  C  would  lie  on  the  opposite  side  of  AB 
from  C. 

Two  plane  triangles  which  are  mutually  equilateral,  and 
consequently  mutually  equiangular,  could  be  made  to  coincide 
by  turning  one  of  them  over  (see  Prop.  IV,  Chap.  I),  but  if  you 
should  turn  a  spherical  triangle  over,  it  would  no  longer  lie 
on  the  sphere. 

Hence,  in  spherical  triangles  whose  parts  are  respectively 
equal,  it  is  necessary  to  distinguish  two  whose  parts  are 
arranged  in  the  same  order  from  two  whose  parts  are  arranged 
in  opposite  orders. 

The  first  are  identically  equal,  the  second  are  symmetrical. 

Proposition  IX  can  then  be  stated  as  follows: 

//  two  spherical  triangles  lying  on  the  same  sphere, 
or  on  equal  spheres,  have  the  three  sides  of  one  equal, 
respectively,  to  the  three  sides  of  the  other,  they  are 
either  identically  equal  or  syrmnetrical,  according  as 
their  parts  are  arranged  in  the  same  order  or  in 
opposite  orders. 

625.  Definition.  If  two  sides  of  a  spherical  triangle  are 
equal,  the  triangle  is  an  isosceles  spherical  triangle. 


Proposition  X 

626.  An  isosceles  spherical  triangle  and  its  symmet- 
rical spherical  triangle  are  identically  equal. 

Compare  Article  463. 

627.  Corollary.     If  two  sides  of  a  spherical  triangle  are 
equal,  the  angles  opposite  those  sides  are  equal. 

Form  its  symmetrical  triangle   on   the  opposite  side   of   the 

sphere  and  superpose  the'm. 
2c 


386  ELEMENTARY  GEOMETRY  [Chap.  IX 


Proposition  XI 

628.  If  two  spherical  triangles  lying  on  the  same 
sphere,  or  on  equal  spheres,  have  two  sides  and  the 
included  angle  of  the  one  equal,  respectively,  to  two  sides 
and  the  included  angle  of  the  other,  they  are  either 
identically  equal  or  symmetrical,  according  as  the  parts 
are  arranged  in  the  same  order  or  in  opposite  orders. 

Compare  Article  466. 

Or,  the  triangles  may  be  superposed  if  the  parts  are  arranged 
in  the  same  order,  or  one  may  be  superposed  to  the  symmetric 
of  the  other  if  the  parts  are  arranged  in  opposite  orders. 


•  Proposition  XII 

629.  //  two  spherical  triangles  lying  on  the  same 
sphere,  or  on  equal  spheres,  have  one  side  and  the  two 
adjacent  angles  of  the  one  equal,  respectively,  to  one  side 
and  the  two  adjacent  angles  of  the  other,  they  are  iden- 
tically equal  or  symmetrical,  according  as  the  parts  are 
arranged  in  the  same  order  or  in  opposite  orders. 

Compare  Article  467,  or  prove  by  superposition. 


EXERCISES 

1.  If  two  angles  of  a  spherical  triangle  are  equal,  the  sides  opposite 
those  angles  are  also  equal. 

2.  The  arcs  of  great  circles  which  bisect  the  angles  of  a  spherical  tri- 
angle are  concurrent. 

3.  The  angle  between  the  planes  of  two  great  circles  has  the  same 
measure  as  the  arc  of  the  great  circle  which  joins  their  nearef  poles. 

4.  If  two  equal  circles  lying  in  different  planes  have  a  common 
diameter,  any  plane  perpendicular  to  that  diameter  intersects  the  circles 
in  four  points  which  lie  upon  a  circle. 


628-630] 


THE  SPHERE 


387 


Section  III 

POLAR  TRIANGLES 

630.  Definition.  If  on  a  sphere  arcs  of  great  circles  are 
described  which  have  the  vertices  of  a  given  spherical  triangle 
for  poles,  another  spherical  triangle  will  be  formed  which  is 
called  the  polar  of  the  given  triangle. 

Let  ABC  be  any  spherical  triangle,  and  let  a',  b\  c'  be  arcs  of 
great  circles  whose  poles  are  A,  B,  C.  If  these  great  circles 
are  fully  drawn,  they  will  divide  the  sphere  into  eight  spherical 


triangles.  That  one,  A'B'C,  of  the  eight  is  called  the  polar 
of  ABC  which  lies  so  that  A  and  A'  are  on  the  same  side  of 
BC',  B  and  B',  on  the  same  side  of  AC;  C  and  C",  on  the  same 
side  of  AB. 

That  the  great  circles  a',  b',  c'  divide  the  sphere  into  eight 
triangles  is  easily  seen  if  it  is  observed  that  b'  and  c'  divide 
the  sphere  into  four  parts,  and  that  a'  divides  each  of  these 
into  two  triangles. 

When  A,  B,  C  are  used  to  denote  the  vertices  of  any  spheri- 
cal triangle,  a,  b,  c  will  be  used  to  denote  the  opposite  sides 
of  that  triangle.  A',  B',  C  to  denote  the  vertices  of  the  polar 
triangle,  a',  b',  c'  to  denote  the  opposite  sides  of  the  polar  tri- 
angle. 

Thus  A,  B,  Care  the  poles  of  the  arcs  a',  6',  c',  and  A',  B',  C 
are  the  poles  of  the  arcs  a,  b,  c. 


388  ELEMENTARY   GEOMETRY  [Chap.  IX 


Proposition  XIII 

631.  //  the  first  of  two  spherical  triangles  is  the  polar 
of  the  second,  then  the  second  is  also  the  polar  of  the 
first. 


Let  the  two  triangles  ABC  and  A'B'C  lie  on  the  same  sphere, 
and  suppose  that  A'B'C  is  the  polar  triangle  of  ABC. 

It  is  required  to  prove  that  ABC  is  also  the  polar  triangle 
of  A'B'C 

Proof.  Because  A  is  the  pole  of  the  arc  B'  C,  B'  is  a  quad- 
rant's distance  from  A.  (Art.  596.) 

And  because  C  is  the  pole  of  A'B',  B'  is  a  quadrant's  dis- 
tance from  C. 

Therefore  B'  is  the  pole  of  the  arc  AC.  (Art.  597.) 

Similarly,  A'  is  the  pole  of  the  arc  BCy  and  C  is  the  pole  of 
the  arc  AB. 

Therefore  A  ABC  is  the  polar  triangle  of  A'B'C 

632.   Definition.     The  supplement  of  any  arc  of  a  circle  is 

that  arc  which  taken  with  it  makes  up  a  semicircle. 

EXERCISES  -^^^    ^^    ^/= 

1.  Every  point  of  the  great  circle  which  bisects  a  given  arc  of  a  ^ea^)     ' 
circle  at  right  angles  is  equidistant  from  the  extremities  of  the  arc.       fh  r    r 

2.  Through  a  given  point  on  a  sphere  describe  an  arc  of  a  great  circle 
making  right  angles  with  a  given  arc  of  a  great  circle. 


631-633] 


THE  SPHERE 


389 


Proposition  XIV 

633.  In  two  polar  triangles,  the  measure  of  any  angle 
in  one  is  equal  to  the  measure  of  the  supplement  of 
that  side  in  the  other  of  which  its  vertex  is  the  pole. 


Let  ABC  and  A'B'C  be  two  polar  triangles  lying  on  any 
sphere,  so  that  A  is  the  pole  of  B'C,  B  is  the  pole  of  A'C, 
and  C  is  the  pole  of  A'B'. 

It  is  required  to  prove  that  the  measure  of  the  angle  A  is 
equal  to  the  measure  of  the  supplement  of  the  arc  B'C,  the 
measure  of  the  angle  B  is  equal  to  the  measure  of  the  supple- 
ment of  the  arc  A'C,  etc. 

Proof.  Produce  the  sides  AB  and  AC,  if  necessary,  to  meet 
the  side  B'C  in  the  points  D  and  E. 

Then  the  measure  of  Z  ^  =  the  measure  of  arc  BE.  (Art.  615.) 

The  arc  B'E  is  a  quadrant.     Why  ? 

For  the  same  reason  DC  is  a  quadrant. 

Therefore  the  sum  of  B'E  and  DC  is  a  semicircle. 

But  the  sum  of  B'E  and  DC  =  the  sum  of  B'E,  DE,  and  EC 

=  the  sum  of  B'C  and  DE. 

Therefore  the  sum  oi  B'C  and  DE  is  a  semicircle,  and  the 
arc  DE  is  the  supplement  of  the  arc  B'C. 

Therefore  the  measure  of  Z  ^  =  the  measure  of  the  supple- 
ment of  the  arc  B'C. 

Similarly  the  other  relations  can  be  proved. 


390  ELEMENTARY  GEOMETRY  [Chap.  IX 

634.  Corollary  I.  If  A  is  any  angle  of  a  spherical  triangle 
a7id  a'  the  corresponding  side  of  its  polar  triangle,  theii  the  meas- 
ure of  a'  is  equal  to  the  measure  of  the  supplement  of  A. 

635.  Corollary  II.  If  any  spherical  triangle  is  equiangu- 
lar, its  polar  triangle  is  equilateral ;  and  conversely. 

636.  Corollary  III.  If  two  spherical  triangles  on  the  same 
sphere,  or  on  equal  spheres,  are  mutually  equiangular,  their  polar 
triangles  are  mutually  equilateral ;  and  conversely. 


Proposition  XV 

637.  The  sum  of  the  angles  of  any  spherical  triangle 
is  greater  than  two,  and  less  than  six,  right  angles. 

First,  let  A,  B,  C  be  the  measures  of  the  angles  of  a  spherical 
triangle,  and  a\  b',  c'  the  measures  of  the  corresponding  sides 
of  its  polar  triangle,  the  units  of  measure  in  the  two  cases  being 
the  angle  degree  and  the  arc  degree,  respectively. 

Then  a'  =  180  -  A.  (Art.  634.) 

b'  =  180  -  B. 

c'  =  180  -  a 

Therefore      a'  -{- b'  -{- c'  =  540  -  (A  +  B -{-  C). 

Now  a'  +  6'  +  c'  is  less  than  360,  the  measure  of  a  great 
circle.  (Prop.  VIII.) 

Therefore  540  -{A  +  B  +  C)i^  less  than  360. 

Hence,  A-\-  B  -{■  C  must  be  greater  than  180,  which  is  the 
measure  of  two  right  angles. 

That  is,  the  sum  of  the  angles  of  the  triangle  is  greater  than 
two  right  angles. 

Next,  a'  4-  6'  4-  c'  must  have  some  value  greater  than  zero. 

Therefore  A -{-  B  -\-  C  \^  less  than  540,  which  is  the  measure 
of  six  right  angles. 


034-641]  THE  SPHERE  391 


Proposition  XVI 

638.  If  two  triangles  on  the  same  sphere,  or  on  equal 
spheres,  are  mutually  equiangular,  they  are  also  mutu- 
ally equilateral,  and  are  either  identically  equal  or  sym- 
metrical, according  as  their  parts  are  arranged  in  the 
same  order  or  in  opposite  orders. 

The  proof  is  left  to  the  pupil.     Apply  Article  636,  and  Proposi- 
tion IX. 

639.  Corollary.  A  spherical  triangle  may  have  one,  two,  or 
three  right  angles,  or  one,  two,  or  three  obtuse  angles. 

Definitions 

640.  A  spherical  triangle  having  two  right  angles  is  called 
a  bi-rectangular  triangle,  and  one  having  three  right  angles  is 
called  a  tri-rectangular  triangle. 

641.  The  excess  of  the  sum  of  the  angles  of  a  spherical  tri- 
angle over  two  right  angles  is  called  the  spherical  excess  of  the 
triangle. 

EXERCISES 

1.  Prove  that  in  a  bi-rectangular  spherical  triangle  the  sides  opposite 
the  right  angles  are  both  quadrants. 

2.  Prove  that  if  two  sides  of  a  spherical  triangle  are  quadrants,  the 
third  side  has  the  same  measure  as  the  opposite  angle. 

3.  Each  side  of  a  tri-rectangular  spherical  triangle  is  a  quadrant. 

4.  If  two  angles  of  a  spherical  triangle  are  supplementary,  the  sum  of 
the  sides  opposite  them  is  equal  to  a  semicircle. 

Suggestion.  Produce  two  sides  of  the  triangle  to  intersect,  thus 
forming  an  adjacent  triangle  having  two  equal  angles. 

5.  The  spherical  excess  of  a  bi-rectangular  spherical  triangle  is  equal 
to  the  measure  of  the  third  angle.  .       ^ 


392  ELEMENTARY  GEOMETRY  [Chap.  IX 


Proposition  XVII 

642.  The  shortest  line  that  can  be  drawn  on  a  sphere 
between  two  points  is  the  arc  of  a  ^reat  circle,  not  greater 
than  a  semicircle,  joining  the  two  points. 


Let  A  and  B  be  any  two  points  on  a  sphere,  and  AFB  the 
arc  of  a  great  circle,  not  greater  than  a  semicircle,  joining 
them ;  also  let  ADCEB  be  any  other  path  on  the  sphere  from 
A  to  B. 

It  is  required  to  prove  that  the  path  AFB  is  shorter  than  the 
path  ADCEB. 

Proof.  In  the  path  ADCEB  take  any  point  C,  and  draw  the 
arcs  of  great  circles  AC  and  CB. 

Then  ACB  is  a  spherical  triangle,  and  the  sum  of  AC  and 
CB  is  greater  than  AB.  (Prop.  VII.) 

In  the  path  ADC  choose  any  point  D,  and  draw  the  arcs  of 
great  circles  AD  and  DC. 

Then  AD  -h  DC  is  greater  than  AC. 

Therefore  AD  +  DC -{-  CB  is  greater  than  AC  +  CB,  which 
is  again  greater  than  AB. 

By  repeating  this  process  indefinitely  the  broken  path  of 
arcs  of  great  circles  increases  in  length  with  each  repetition, 
and  so  differs  more  and  more  from  the  path  AFB. 

Also  the  broken  path  of  arcs  of  great  circles  can  eventually 
be  made  to  differ  as  little  as  we  please  from  the  given  path 
ADCEB. 

Therefore  the  path  AFB  is  shorter  than  the  path  ADCEB. 


642-643] 


THE  SPHERE 


393 


Proposition  XVIII 

643.   To  find  the  length  of  the  diameter  of  a  given 
material  sphere. 


Let  PABC  be  a  given  material  sphere. 

It  is  required  to  find  tlie  lengtti  of  its  diameter. 

Construction.  With  any  point  P  on  the  given  sphere  as 
pole  describe  a  circle  ABC. 

With  the  compasses  take  the  lengths  of  the  chords  AB,  BC, 
CA,  and  construct  in  any  plane  a  triangle  A'B'C  whose  sides 
are  equal  to  these  three  lengths.  About  A'B'C  describe  a 
circle  which  will  be  equal  to  the  circle  ABC.  Find  its  centre 
Q'  and  the  length  of  its  radius  B'Q'  =  BQ 

With  the  compasses  take  the  length  of  the  chord  BP  and 
construct  a  right  triangle  pbq,  having  the  hypotenuse  b}^  —  BP 
and  the  side  6g  =  BQ. 

Draw  hp  perpendicular  to  &p,  meeting  ipq  produced  at  p'. 

Then  /?p'  is  equal  in  length  to  a  diameter  of  the  given  sphere. 

The  proof  is  left  to  the  pupil. 


EXERCISES 

1.  Three  great  circles  of  a  sphere  which  do  not  intersect  in  a  common 
point,  divide  the  surface  of  the  sphere  into  four  pairs  of  symmetrical 
triangles. 

2.  A  tri-rectangular  spherical  triangle  coincides  with  its  polar  triangle. 


394  ELEMENTARY   GEOMETRY  [Chap.  IX 

Section  IV 

AREAS   OF   SPHERICAL   TRIANGLES 

644.  Definition.     The  area  of  a  spherical  triangle  is  that 
portion  of  the  sphere  which  is  enclosed  by  the  triangle. 

645.  Axiom  15.     Two  spherical  triangles  which  are  identically 
equal  have  equal  areas. 

Proposition  XIX 

646.  Two  symmetrical  spherical  triangles  have  equal 
areas. 


Let  yl^Cand  A'B'C  be  two  symmetrical  spherical  triangles 
lying  on  the  same  sphere. 

It  is  required  to  prove  that  ABC  and  A'B'C  have  equal  areas. 

Proof.  Two  symmetrical  spherical  triangles  can  always  be 
placed  so  as  to  be  vertically  opposite  on  the  sphere.     Why  ? 

Assume  that  the  given  triangles  are  so  placed. 

Let  P  be  the  pole  of  the  small  circle  through  A,  B,  C,  and 
draw  the  diameter  POP'. 

Draw  also  the  arcs  of  great  circles  PA^  PB,  PC,  PA',  PB\ 
PC. 

Now  PA  =  PA',  PB  =  PB',  PC  =  P'C,     (Art.  157.) 

also  PA=  PB==  PC.  (Prop.  II.) 

Therefore  PA'  =  PB'  =  PC 


644-648]  THE  SPHERE  395 

As  FAB  and  P'A'B'  are  symmetrical  triangles  on  the  sphere, 
and  they  are  also  isosceles  triangles. 

Therefore  As  PAB  and  PA'B'  are  identically  equal  and 
have  equal  areas.  (Prop.  X.) 

Similarly  As  PBC  and  P'B'C  have  equal  areas. 

Also  As  PAC  and  PA'C  have  equal  areas. 

Therefore,  adding.  As  ABC  and  A'B'C  have  equal  areas. 

If  the  pole  P  falls  outside  of  A  ABC,  instead  of  adding  all 
three  isosceles  triangles,  two  should  be  added  and  the  third 
subtracted  from  their  sum,  in  order  that  the  result  may  give 
A  ABC. 

Note.  The  triangles  mentioned  in  Propositions  IX,  XI,  and  XII, 
as  being  either  identically  equal  or  symmetrical,  are  in  every  case  equal 


Definitions 

647.  Two  great  circles  on  a  sphere  divide  the  sphere  into 
four  parts,  each  of  which  is  called  a  lune. 

A  lune  then  is  that  portion  of  a 
sphere  enclosed  by  two  halves  of  great 
circles. 

The  spherical  angle  made  by  the  semi- 
circles is  called  the  angle  of  the  lune. 

Of  the  four  lunes  formed  by  two  great 
circles  the  opposite  ones  have  equal 
angles. 

Opposite  lunes  on  a  sphere  are  equal,  since  they  can  be 
superposed. 

648,  Theorem.     Any  two  lunes,  having  equal  angles,  on  the 
same  sphere  or  on  equal  spheres  are  identically  equal. 

Of  two  lunes  having  unequal  angles,  on  the  same  sphere  or 
on  equal  spheres,  that  which  has  the  greater  angle  is  the 
greater ;  and  of  two  lunes  having  equal  angles  on  unequal 
spheres,  that  which  lies  on  the  greater  sphere  is  the  greater. 


396  ELEMENTARY  GEOMETRY  [Chap.  IX 


Proposition  XX 

649.   The  ratio  of  two  lunes  on  the  same  sphere  or  on 
equal  spheres  is  equal  to  the  ratio  of  their  angles. 


Let  PAP'B  and  PBFC  be  two  lunes  on  the  same  sphere 
whose  angles  are  APB  and  BPC. 

It  is  required  to  prove  that 

lune  PAPB  :  lune  PBP'C  =  Z  APB  :  Z  BPC. 

Outline  of  proof.     There  are  two  cases  to  be  considered. 

First,  when  ZsAPB  and  BPC  are  commensurable. 

Take  a  common  measure  of  the  two  angles  and  divide  them 
into  equal  parts. 

By  planes  through  the  centre  of  the  sphere  divide  the  lunes 
into  parts  having  equal  angles,  and  hence  equal. 

Next,  when  Zs  APB  and  BPC  are  incommensurable. 

Take  a  measure  of  one  angle  and  apply  it  as  often  as  possible 
to  the  other.  Then  proceed  as  above,  and  finally  make  the 
unit  of  measure  indefinitely  small. 

As  a  model  take  the  proof  of  Prop.  X  in  Chap,  III. 

650.  Corollary.  The  area  of  a  lune  is  to  the  area  of  the 
whole  sphere  in  the  same  ratio  as  the  angle  of  the  lune  is  to  four 
right  angles. 


649-652] 


THE  SPHERE 


397 


Proposition  XXI 

651.  //  two  great  circles  intersect  on  a  hemisphere,  the 
areas  of  the  two  triangles  formed  by  their  arcs  and  arcs 
of  the  great  circle  hounding  the  hemisphere  are  together 
equal  to  a  lune  having  the  same  angle  as  the  angle 
between  the  great  circles. 


Let  ABA'B'  be  the  great  circle  of  the  hemisphere  PABA'B', 
and  let  the  two  great  circles  APA'  and  BPB'  intersect  at  P. 

It  is  required  to  prove  that  the  triangles  APB  and  A'PB'  are 
together  equal  to  the  lune  PAP'B  whose  angle  is  APB. 

Proof.     A  PA'B'  is  symmetrical  with  A  PAB. 

Hence theareaofAP^'J5'=theareaofAP'^jB.  (Prop  XIX.) 

But  A  PAB  and  A  PAB  make  up  the  lune  PAPB. 

Therefore  A  PAB  and  A  PA'B'  are  together  equal  in  area  to 
the  lune  PAPB. 

SPHERICAL   UNITS 

652.  The  ordinary  unit  of  surface  is  the  area  of  a  plane 
square  whose  dimensions  are  of  unit  length,  but  a  different 
unit    is    sometimes    found    convenient  p 

in  the  measurement  of  spherical  areas. 
This  unit  we  shall  call  a  spherical  unit 
of  surface. 

Definition.  A  spherical  unit  of  sur- 
face is  the  area  of  a  bi-rectangular 
spherical  triangle  whose  third  angle  is 
one  degree. 


398  ELEMENTARY  GEOMETRY  [Chap.  IX 

In  other  words,  the  spherical  unit  of  surface  is  a  half  lune 
whose  angle  is  one  degree,  the  bisection  being  made  by  an  arc 
of  a  great  circle  perpendicular  to  both  arcs  of  the  lune. 

It  should  be  carefully  noticed  that  the  spherical  unit  of 
surface  is  different  on  different  spheres. 

The  area  of  a  hemisphere  is  360  spherical  units;  of  a  sphere  is  720 
spherical  units. 

The  area  of  a  bi-rectangular  triangle  expressed  in  spherical  units  is 
equal  to  the  measure  of  its  third  angle. 

The  area  of  a  lune  expressed  in  spherical  units  is  equal  to  twice  the 
measure  of  its  angle. 

Proposition  XXII 

653.  The  area  of  any  spherical  triarigle  expressed  in 
spherical  units  is  equal  to  the  spherical  excess  of  the 
triangle. 

Q 


G 

Let  ABC  be  any  spherical  triangle  lying  on  the  hemisphere 
ABCPQ. 

It  is  required  to  prove- that  the  area  of  ABC  expressed  in 
spherical  units  equals  the  excess  of  the  sum  of  the  angles  A, 
B,  C,  over  two  right  angles. 

Proof.  The  hemisphere  consists  of  the  four  triangles  ABC, 
ACP,  APQ,  AQB. 

A  ABC  +  A  APQ  =  lune  whose  angle  is  A.    (Prop.  XXI.) 
A  ABC  +  A  ACP  =  lune  whose  angle  is  B. 
A  ABC  +  A  AQB  =  lune  whose  angle  is  C. 


652-654]  THE  SPHERE 

Adding  the  areas  we  get : 

Twice  area  of  A  ABC  +  area  of  hemisphere 

=  area  of  lune  whose  angle  is  J.  +  area  of  lune  whose  angle  is  B 

+  area  of  lune  whose  angle  is  C. 

Now  the  area  of  a  hemisphere  equals  360  spherical  units, 
and  the  area  of  the  lune  whose  angle  is  A  equals  2 A  spherical 
units.     Therefore,  (Art.  652.) 

Twice  area  of  A^^C  +  360  spherical  units  =  (2  ^  +  2  i3 
+  2  C)  spherical  units.     Or, 

Area  of  A  ABC  -f  180  spherical  units  =  (A  -{-  B  -{-  C)  spheri- 
cal units. 

Therefore  area  of  A  ABC  =  (A -]- B  +  C  -  180)  spherical 
units. 

But  A-{-B-{-  O— 180  is  the  spherical  excess  of  the  triangle. 

Therefore  area  of  A  ABC  =  as  many  spherical  units  as  repre- 
sents its  spherical  excess. 

654.  Corollary.  If  S  is  the  sum  of  the  angles  of  a  spherical 
polygon  of  n  sides,  the  area  of  the  polygon  is  [«S  —  (n  —  2)  180] 
spherical  units. 

Divide  the  polygon  into  triangles  by  passing  arcs  of  great  circles 
tiirougli  one  vertex. 

EXERCISES 

1.  If  three  great  circles  are  so  drawn  on  a  sphere  that  each  is  perpen- 
dicular to  the  other  two,  into  what  sort  of  triangles  do  they  divide  the 
sphere  ?     What  are  their  areas  expressed  in  spherical  units  ? 

2.  If  the  sides  of  a  spherical  triangle  are  75,  120,  and  95  arc  degrees, 
respectively,  what  is  the  measure  of  the  angles  of  its  polar  triangle  ? 
What  is  the  area  of  the  polar  triangle  ? 

3.  If  the  angles  of  a  spherical  triangle  are  85,  1.35,  and  65  degrees, 
respectively,  what  is  its  area,  and  what  are  the  lengths  of  the  sides  of  its 
polar  triangle  ? 

4.  The  sum  of  the  interior  angles  of  a  convex  spherical  hexagon  is 
greater  than  eight  and  less  than  twelve  right  angles. 


400 


ELEMENTARY  GEOMETRY 


[Chap.  IX 


Proposition  XXIII 

655.  The  area  of  the  surface  generated  by  a  line-seg- 
merit  revolving  about  an  axis  in  its  plane  is  equal  to 
the  length  of  the  projection  of  the  line-segment  on  the 
axis  multiplied  by  the  circuinference  of  the  circle  whose 
radius  is  equal  to  that  segment  of  the  perpendicular 
bisector  of  the  revolving  line-segmjent  which  is  inter- 
cepted between  it  and  the  axis. 


Let  PQ  be  any  given  line-segment  which  revolves  about  an 
axis  a  lying  in  a  plane  with  it ;  let  pq  be  the  projection  of  PQ 
on  a,  and  RS  be  the  perpendicular  bisector  of  PQ  intercepted 
between  PQ  and  a. 

It  is  required  to  prove  that  the  area  of  the  surface  generated 
by  PQ  is  equal  to  pq  multiplied  by  the  circumference  of  the 
circle  whose  radius  is  RS. 

Proof.  Draw  Rr  perpendicular  to  a,  and  PT  parallel  to  a. 
As  PQ  revolves  about  a  it  generates  the  lateral  surface  of  the 
frustum  of  a  right  circular  cone. 

The  area  of  this  surface  =  PQ  x  circ.  of  circle  of  radius  Rr. 

(Ex.  9,  p.  367.) 

Now  A  PTQ  and  RrS  are  similar.     Why  ? 

Therefore,  since  PT  —  pq, 


PQ^RS^  circ.  of  circle  of  rad.  RS 
pq       Rr      circ.  of  circle  of  rad.  Rr 


(Art.  362.) 


655-656] 


THE  SPHERE 


401 


Hence  PQ  x  circ.  of  circle  of  rad.  Rr=pq  x  circ.  of  circle 
of  rad.  MS. 

Therefore  area  of  surface  generated  by  PQ  =  j)^  x  circ.  of 
circle  of  rad.  RS. 

If  PQ  meets  a  at  P,  the  surface  generated  is  conical,  but 
the  proposition  and  proof  still  hold,  as  is  easily  shown. 
Also,  if  PQ  is  parallel  to  a  the  proposition  holds. 

What  changes  are  necessary  in  the  statement  of  the  theorem 
and  in  the  proof  if  PQ  intersects  a  ? 

Note.  The  pupil  should  be  careful  to  observe  that  in  this  proposition 
the  area  spoken  of  is  the  measure  of  the  surface  expressed  in  ordinary 
plane  units,  so  many  square  feet,  or  square  inches,  not  in  spherical  units. 


Definitions 

656.   A  segment  of   a  sphere,  or  a  spherical  segment,  is  the 

figure  formed  by  the  planes  of  two  parallel  sections  of  the  sphere 
and  the  spherical  surface  intercepted  be- 
tween them. 

The  plane  faces  of  a  segment  are  called 
its  bases,  and  the  spherical  surface  is  called 
a  zone. 

Hence,  a  zone  is  a  portion  of  a  sphere 
intercepted  between  two  parallel  planes 
which  cut  the  sphere. 

If  one  of  the  parallel  planes  forming 
the  bases  of  a  spherical  segment  is  tan- 
gent to  the  sphere,  the  segment  is  a 
segment  of  one  base,  and  its  spherical  sur- 
face is  a  zone  of  one  base. 

The    perpendicular    distance   between 
the  bases  of  a  spherical  segment  is  the 
altitude  of  the  segment.      It  is  also  the 
altitude  of  the  zone. 
2d 


Spherical  segment 


Segment  of  base 


402  ELEMENTARY  GEOMETRY  [Chap.  IX 


Proposition  XXIV 

657.    The  area  of  a  zone  is  equal  to  the  product  of  its 
altitude  and  the  circumference  of  a  great  circle. 


Let  AB  be  an  arc  of  a  great  circle  of  a  sphere  whose  centre 
is  0  and  diameter  PP' -^  and  suppose  AB  to  rotate  about  the 
diameter  PP  so  as  to  generate  a  zone  whose  altitude  is  ah. 

It  is  required  to  prove  that  the  area  of  the  zone  generated  by 
AB  is  equal  to  the  product  of  ah  and  the  circumference  of  a 
great  circle  of  the  sphere. 

Proof.  Draw  the  chord  AB  and  the  perpendicular  bisector 
EO,  which  passes  through  the  centre  of  the  sphere. 

The  area  of  the  surface  generated  by  the  chord  AB  as  it 
rotates  about  PP  =  ah  X  circ.  of  a  circle  of  rad.  OE. 

(Prop.  XXIII.) 

Let  the  arc  AB  he  bisected  at  C,  and  draw  the  chords  AC 
and  GB.     Also  the  perpendicular  bisectors  OF  and  OH. 

Let  c  be  the  projection  of  C  upon  PP. 

Then  the  area  of  the  surface  generated  by  the  two  chords 
AC  and  CB  as  they  rotate  about  PP'  =  ac  x  circ.  of  circle  of 
rad.  OF  +  c6  X  circ.  of  circle  of  rad.  OH 

=  ah  X  circ.  of  circle  of  rad.  OF,  since  0H=  OF    Why  ? 


657-660]  THE  SPHERE  403 

If  this  process  of  bisecting  the  arcs  is  repeated  indefinitely, 
the  series  of  chords  approaches  the  arc  AB  as  a  limit,  and  the 
perpendicular  bisector  approaches  a  radius  r  of  the  sphere  as  a 
limit. 

Therefore  the  area  of  the  surface  generated  by  the  arc  AB 
as  it  rotates  about  PP  —  ah  x  circ.  of  circle  of  rad.  r. 

That  is,  the  area  of  a  zone  whose  altitude  is  ab  =  ab  x  circ. 
of  a  great  circle. 

658.  Corollary  I.  Hie  area  of  a  zone  whose  altitude  is  h, 
lying  on  a  sphere  of  radius  r,  equals  2  irrh. 

659.  Definition.  A  right  circular  cylinder  the  radius  of 
whose  base  equals  the  radius  of  a  sphere,  and  whose  altitude 
equals  the  diameter  of  the  sphere,  when  placed  so  as  to  enclose 
the  sphere,  is  called  the  enveloping  cylinder  of  the  sphere. 

660.  Corollary  II.  The  area  of  a  zone  is  equal  to  that 
portion  of  the  lateral  area  of  the  enveloping  cylinder  which  is  inter- 
cepted between  the  same  planes  as  the  zone. 

For  the  circumference  of  the  base  of  the  enveloping  cylinder 
is  2  7rr,  and  if  h  is  the  distance  between  the  planes  of  section, 
the  lateral  area  of  the  cylinder  intercepted  between  them  is 
2  irrh  (Art.  561)  and  this  is  also  the  area  of  the  intercepted 
zone. 

EXERCISES 

1.  Find  the  area  of  a  zone  of  altitude  3  inches  on  a  sphere  whose 
radius  is  10  inches. 

2.  The  diameter  of  a  given  sphere  is  10  inches  and  it  is  desired  to  cut 
the  surface  by  parallel  planes,  into  five  parts  of  equal  area.  Locate  the 
planes  of  section. 

3.  A  slice  is  cut  from  a  sphere  of  radius  5  inches  by  a  plane  passing 
3  inches  from  the  centre ;  what  is  the  area  of  the  spherical  surface,  and 
also  of  the  plane  surface  of  the  slice  ? 

4.  Prove  that  the  area  of  a  zone  of  one  base  is  equal  to  the  area  of  a 
circle  whose  radius  is  the  chord  of  the  generating  arc  of  the  zone. 


404  ELEMENTARY  GEOMETRY  [Chap.  IX 


Proposition  XXV 

661.  The  area  of  a  sphere  is  equal  to  the  product  of  a 
diameter  and  the  circumference  of  a  great  circle. 

This  follows  directly  from  Proposition  XXIV,  the  generating 
arc  being  a  semicircle. 

662.  CoROLLAKY  I.  Tlie  area  of  a  sphere  ivJiose  radius  is  r 
is  given  by  the  formula  : 

A  =  2rx27rr=:4:7rr'. 

663.  Corollary  II.  The  areas  of  two  spheres  are  in  the 
same  ratio  as  the  squares  of  their  radii,  or  as  the  squares  of  their 
diameters. 

664.  Corollary  III.  Tlie  area  of  a  sphere  equals  four 
times  the  area  of  one  of  its  great  circles. 

665.  Corollary  IV.  Tlie  area  of  a  sphere  equals  the  lateral 
area  of  its  enveloping  cylinder. 

666.  Corollary  Y.     The  area  of  a  spherical  unit  expressed 

in  plane  units  equals 

^  180 

Note.     All  the  above  areas  are  expressed  in  plane  units. 

EXERCISES 

1.  What  is  the  area  of  a  sphere  whose  radius  is  10  inches  ? 

2.  A  sphere  has  a  radius  of  9  inches.  What  is  the  radius  of  a  sphere 
having  double  the  area  ? 

3.  Two  parallel  planes  cut  a  sphere  of  radius  10  inches,  6  and  8  inches, 
respectively,  from  the  centre.  Find  the  area  of  the  zone  so  formed  (1) 
when  the  planes  are  on  the  same  side  of  the  centre,  (2)  when  the  planes 
are  on  opposite  sides  of  the  centre. 

4.  A  triangle  whose  angles  are  80°,  120°,  135°  lies  on  a  sphere  of  radius 
16  inches.     Find  its  area. 

Suggestion.  First  find  the  area  in  spherical  units  and  then  reduce  to 
plane  units. 


661-669]  THE  SPHERE  405 

Section  V 

VOLUME   OF  THE   SPHERE 
Proposition  XXVI 

667.  The  volume  of  a  sphere  is  equal  to  one-third  the 
product  of  its  area  and  its  radius. 

Suppose  that  a  polyhedron  of  any  number  of  sides  has  been 
circumscribed  about  a  sphere.  Through  all  the  edges  of  the 
polyhedron  and  the  centre  of  the  sphere  pass  planes. 

This  divides  the  polyhedron  into  as  many  pyramids  as  the 
polyhedron  has  faces.  The  altitude  of  each  pyramid  is  equal 
to  a  radius  of  the  sphere,  since  each  face  of  the  polyhedron  is 
tangent  to  the  sphere. 

The  volume  of  any  one  of  the  pyramids  =  i  r  x  area  of  its 
base  (Art.  523),  and  the  sum  of  all  the  bases  of  the  pyramids 
equals  the  surface  of  the  polyhedron. 

Therefore,  volume  of  the  polyhedron  =  |  r  x  the  surface  area 
of  the  polyhedron,  and  this  relation  holds  no  matter  how  many 
faces  the  polyhedron  may  have. 

Now  by  increasing  the  number  of  faces  of  the  circumscribed 
polyhedron  indefinitely,  it  can  be  made  to  approach  the  sphere 
as  its  limit,  and  its  volume  will  approach  the  volume  of  the 
sphere  as  its  limit. 

Therefore,  volume  of  a  sphere  =  J  r  x  area  of  the  sphere. 

668.  Corollary  I.  The  volume  of  a  sphere  can  be  expressed 
by  the  following  formula: 

F=  1  r  X  4  TT?^  =  1 7r7-3.  (Art.  662.) 

669.  Corollary  II.     The  volumes  of  two  spheres  are  in  the 

same  ratio  as  the  cubes  of  their  radii,  or  as  the  cubes  of  their 

diameters. 

EXERCISES 

1.  Show  that  the  volume  of  a  sphere  is  equal  to  two-thirds  of  the 
volume  of  the  enveloping  cylinder. 


406 


ELEMENTARY  GEOMETRY 


[Chap.  IX 


Spherical  pyramid 


Definitions 

670.  A  spherical  pyramid  is  a  figure  consisting  of  a  spherical 
polygon  for  base  and  plane  faces  through 
the  sides  of  the  polygon. 

The  lateral  edges  of  the  pyramid  are 
all  radii  of  the  sphere. 

671.  A  spherical  sector  is  a  figure 
consisting  of  a  zone  and  two  conical  sur- 
faces, having  the  centre  of  the  sphere 
for  vertex,  and  passing  through  the  cir- 
cular boundaries  of  the  zone. 

If  the  zone  is  a  zone  of  one  base,  then  the 
spherical  sector  has  but  one  conical  surface. 

The  zone  is  called  the  base  of  the 
spherical  sector. 

If  a  sector  of  a  circle  is  rotated  about 
any  diameter  of  the  circle,  it  will  gener- 
ate a  spherical  sector.  If  it  is  rotated 
about  one  of  its  own  radii,  the  base  of  Spherical  sector 

the  spherical  sector  will  be  a  zone  of  one  base. 

672.  Theorem.  Tlie  volume  of  a  spherical  pyramid  or  a 
spherical  sector  is  to  the  volume  of  its  sp)here  in  the  same  ratio 
as  the  area  of  its  base  is  to  the  area  of  the  sphere. 

673.  If  V  represents  the  volume  of  a  spherical  pyramid  or  sector,  V 
the  volume  of  the  sphere,  A  the  area  of  the  base  of  the  pyramid  or  sector, 
and  A'  the  area  of  the  sphere, 


whence  F=  ^1  • -' =  J  •  i^  =  i  ^r. 

A'  4  7rr2      ^ 

That  is  to  say  : 

The  volume  of  a  spherical  pyramid,  or  of  a  spherical  sector,  is 
equal  to  one-third  the  product  of  the  area  of  its  base  arid  the 
radius  of  the  sphere. 


670-674] 


THE  SPHERE 


407 


Proposition  XXVII 
674.  To  find  the  volume  of  a  spherical  segment. 


First,  to  find  the  volume  of  a  spherical  segment  of  one  base. 

Suggestions.  Find  the  area  of  the  zone  G-ABC.  (Art.  657.) 
Find  the  volume  of  the  spherical  sector  0-ABCO.  (Art.  673.) 
Subtract  the  volume  of  the  cone  0-ABC.  (Art.  579.) 

Hence  volume  of  the  segment  G-ABC 

where  r  =  radius  of  the  sphere,  a  =  radius  of  base  of  the  seg- 
ment, h  =  altitude  of  the  segment. 


(Art.  255.) 


Then 

V=-[2i''h-ah-{-a%]. 
o 

Now 

h(2  r-h)  =  a\ 

(A 

Hence 

2r'h  =  ah  +  hh. 

Therefore 

F=|[^V  +  a2^] 

=  ^  ira^h  +  \  irh^,  since  r  ■■ 

a'  +  h^ 

2h 

Next,  to  find  the  volume  of  a  spherical  segment  of  two  bases 
lying  on  the  same  side  of  the  centre  of  the  sphere. 

Let  a  and  b  be  the  radii  of  the  two  bases,  a  being  the  greater, 
and  III  the  altitude  of  a  segment  of  one  base  of  radius  a,  h^  the 


408  ELEMENTARY  GEOMETRY  [Chap.  IX 

altitude  of  a  segment  of  one  base  of  radius  h,  so  that  hi  —  hz  =  h, 
the  altitude  of  the  given  segment. 

Then  the  volume  of  the  given  segment 

=  [i  /^l  •  TT  a^  +  I  TT  7^1^  -   [I-  7,2  .  TT  6^  +  i  TT  ^/] 

=  ^[^1^'  +  hb'  -  ha'  -  hjf]  + 1  [7i/  _  /i/  +  3  h^a?  -  3  hyh'''] 
2i  6 

having  added  and  subtracted  li^'  and  li<fi'. 

Now  7ii(2  r  -  7ii)  =  a\  or  2  r7ii  -  7^1^  =  a^ ; 

7i2(2  r  -  7i2)  =  6^  or  2  rU^  -  hi  =  61 

Multiplying  the  first  line  by  3  h^,  the  second  by  3  h^^  and 
subtracting,  gives 

3  h^i'  -  3  Ai62  =  3  fi^i  _  3  7ij2^2. 
Therefore 

^  o 

This  result  may  be  stated  in  words  as  follows : 

The  volume  of  a  spherical  segment  is  equal  to  the  product  of  its 
altitude  and  half  the  sum  of  the  areas  of  its  bases  together  with 
the  volume  of  a  sphere  whose  diameter  is  equal  to  the  altitude  of 
the  segment. 

The  algebraic  difficulties  in  the  reduction  are  only  intro- 
duced for  the  sake  of  giving  the  result  in  a  convenient  form. 
It  should  be  noticed  that  the  last  formula  for  the  volume  will 
reduce  to  that  given  for  a  segment  of  one  base  by  putting  &  =  0. 


674]  THE  SPHERE  409 

MISCELLANEOUS    EXERCISES 

Note.    Use  3^  as  the  approximate  value  of  tt. 

1.  Find  the  area  and  the  volume  of  a  sphere  whose  radius  is  10  inches. 

2.  How  many  square  inches  in  the  spherical  unit  on  a  sphere  whose 
diameter  is  14  inches  ? 

3.  Find  in  square  inches  the  area  of  a  spherical  triangle  the  sum  of 
whose  angles  is  210  degrees,  on  a  sphere  whose  radius  is  15  inches. 

4.  On  a  sphere  of  radius  12  inches,  there  is  described  a  spherical  tri- 
angle and  a  spherical  quadrilateral,  the  sum  of  each  of  whose  angles  is 
450  degrees.     What  is  the  difference  in  their  areas  ? 

5.  Find  the  area  of  a  zone  of  height  3  inches  on  a  sphere  of  radius  10 
inches.    Does  it  matter  where  the  zone  is  placed  on  the  sphere  ? 

6.  The  radii  of  two  spheres  are  in  the  ratio  of  2 : 3.  Compare  their 
areas  and  their  volumes. 

7.  The  area  of  a  sphere  is  1000  square  inches  ;  what  is  the  area  of  one 
of  its  great  circles  ? 

8.  Two  spheres  of  lead,  of  radius  2  inches  and  3  inches,  respectively, 
are  melted  into  a  single  sphere.     Find  its  radius. 

9.  A  sphere  is  such  that  its  number  of  units  of  volume  equals  twice 
its  number  of  units  of  area.     Find  its  radius. 

10.  The  base  of  a  spherical  pyramid  is  a  pentagon  the  sum  of  whose 
angles  is  900  degrees.  Find  the  volume  of  the  pyramid  if  the  radius  of 
the  sphere  is  12  inches. 

11.  If  the  area  of  a  zone  of  one  base  is  one-quarter  of  the  area  of  the 
isphere,  show  that  the  altitude  of  the  zone  is  half  the  radius  of  the  sphere. 

12.  How  far  must  the  eye  be  from  a  sphere  of  radius  16  inches  in  order 
to  see  one-quarter  of  it  ? 

13.  If  the  radius  of  the  base  of  a  circular  cone  and  of  a  circular  cylinder 
equals  the  radius  of  a  sphere  and  they  are  all  of  the  same  height,  show 
tiiat  the  volumes  of  the  cone  and  sphere  are  together  equal  to  the  volume 
of  the  cylinder. 

14.  The  radii  of  the  bases  of  a  spherical  segment  are  6  and  8  inches, 
respectively,  and  its  altitude  is  2  inches.  Find  its  volume,  the  radius  of 
the  sphere  from  which  it  was  cut,  the  areas  of  its  bases,  and  its  lateral 
area. 

15.  Find  the  volume  of  a  sphere  whose  radius  is  14  feet. 

16.  The  volume  of  a  sphere  is  10286  cubic  inches.    Find  its  area. 


410  ELEMENTARY  GEOMETRY  [Chap.  IX 

SUMMARY   OF   CHAPTER  IX 
1.  Definitions. 

(1)  Sphere  —  a  closed  surface  such  that  all  points  of  it  are  equidistant 

from  a  fixed  point  within  it.     §  680. 

(2)  Centre,  Radius,  Diameter.     §  580. 

(3)  Concentric  Spheres  —  two  spheres  having  the  same  centre,  but  not 

coinciding.     §  582. 

(4)  Great  Circle  of  a  Sphere  —  a  circle  lying  on  the  sphere,  whose 

plane  passes  through  the  centre  of  the  sphere.    §  585. 

(5)  Quadrant  —  one-quarter  of  a  great  circle.     §  585. 

(6)  Axis  of  a  Circle  of  a  Sphere  —  the  diameter  of  the  sphere  perpen- 

dicular to  the  plane  of  the  circle.     §  592. 

(7)  Poles  of  a  Circle  of  a  Sphere  —  the  extremities  of  the  axis  of  the 

circle.     §  592. 

(8)  Tangent  to  a  Sphere  —  a  line  or  plane  which  meets  the  sphere  at 

one,  and  only  one,  point.     §  600. 

(9)  Inscribed   Polyhedron  —  one  whose  vertices  lie  on  the  sphere 

§  606. 

(10)  Circumscribed  Polyhedron  —  one  whose  faces  are  tangent  to  the 

sphere.     §  606. 

(11)  Spherical  Angle  —  the  angle  formed  by  two  intersecting  arcs  of 

great  circles,  i.e.  by  the  tangents  at  their  common  point.    §  612. 

(12)  Spherical  Polygon  —  a  closed  figure  on  a  sphere  consisting  of  arcs 

of  great  circles  which  intersect,  two  and  two,  in  order.     §  617. 

(13)  Spherical  Excess  of  a  Triangle  —  the  excess  of  the  sum  of  the 

angles  of  a  spherical  triangle  over  two  right  angles.     §  641. 

(14)  Area  of  a  Spherical  Triangle  —  that  portion  of  the  sphere  which 

is  enclosed  by  the  triangle,  or  the  measure  of  that  portion. 
§644. 

(15)  Lune  —  a  portion  of  a  sphere  enclosed  by  two  halves  of  great 

circles.     §  647. 

(16)  Angle  of  a  Lune  —  the  angle  between  its  boundaries.    §  647. 

(17)  Spherical  Unit  of  Surface  —  the  area  of  a  bi-rectangular  spherical 

triangle  whose  third  angle  is  one  degree.     §  652. 

(18)  Spherical  Segment  —  the  figure  formed  by  the  planes  of  two 

parallel  sections  of  a  sphere  and  the  spherical  surface  inter- 
cepted between  them.    §  656. 

(19)  Zone  —  the  portion  of  a  sphere  intercepted  between  two  parallel 

planes  which  cut  the  sphere.     §  656. 

(20)  Altitude  of  a  Spherical  Segment  or  Zone  —  the  distance  between 

its  bases.    §  666. 


Summary]  THE  SPHERE  411 

(21)  Spherical  Pyramid  —  a  figure  consisting  of  a  spherical  polygon 

for  base,  and  plane  faces  through  the  sides  of  the  polygon. 
§  670. 

(22)  Spherical  Sector  —  a  figure  consisting  of  a  zone  and  two  conical 

surfaces  through  the  circular  boundaries  of  the  zone,  having  the 
centre  of  the  sphere  for  vertex.     §  671. 

(23)  Enveloping  cylinder  of  a  sphere  —  a  right  circular  cylinder  the 

radius  of  whose  base  equals  the  radius  of  the  sphere,  and 
whose  altitude  equals  the  diameter  of  the  sphere,  placed  so 
as  to  enclose  the  sphere.    §  659. 

2.  Axioms. 

(1)  Two  spherical  triangles  which  are  identically  equal  have  equal 
areas  (Axiom  15).     §  645. 

3.  Problems. 

(1)  To  find  the  length  of  the  diameter  of  a  given  material  sphere. 
§643. 

4.  Theorems  on  the  Properties  of  a  Sphere  and  its  Plane  Sections. 

(1)  Equal  spheres  have  equal  radii  and  equal  diameters,  and  con- 

versely.    §  581. 

(2)  The  section  of  a  sphere  made  by  a  plane  is  a  circle.     §  583. 

(3)  Sections  of  a  sphere  made  by  planes  equidistant  from  the  centre 

are  equal  circles,  and  of  two  sections  made  by  planes  unequally 
distant  from  the  centre,  that  is  the  greater  circle  which  is  made 
by  the  nearer  plane.     §  584. 

(4)  The  centre  of  any  circle  of  a  sphere  is  the  foot  of  the  perpen- 

dicular drawn  from  the  centre  of  the  spl.ere  to  the  plane  of  the 
circle.    §  586. 

(5)  The  centres  of  all  great  circles  coincide  with  the  centre  of  the 

sphere,  and  all  great  circles  on  the  same  sphere  are  equal. 
§587. 

(6)  Every  great  circle  divides  the  sphere  into  two  equal  parts.    §  588. 

(7)  Any  two  great  circles  on  the  same  sphere  bisect  each  other. 

§589. 

(8)  An  arc  of  a  great  circle  can  be  drawn  through  any  two  given 

points  on  a  sphere,  and  if  the  two  given  points  are  not  the 
extremities  of  a  diameter,  only  one  such  arc  less  than  a  semi- 
circle can  be  drawn.    S  590. 


412  ELEMENTARY  GEOMETRY  [Chap.  IX 

(9)  Through  three  given  points  on  a  sphere,  one  and  only  one  circle 
can  be  drawn.     §  591. 

(10)  All  points  of  any  circle  of  a  sphere  are  equidistant  from  either  of 

Its  poles.    §  594. 

(11)  The  polar  distance  of  any  point  of  a  great  circle  is  a  quadrant. 

§596. 

(12)  If  a  point  on  a  sphere  is  a  quadrant's  distance  from  each  of  two 

given  points  on  the  sphere,  it  is  the  pole  of  the  great  circle 
passing  through  these  two  points.     §  597. 

(13)  If  one  point  on  a  sphere  is  a  quadrant's  distance  from  another, 

it  is  the  pole  of  some  great  circle  passing  through  the  other. 
§598. 

(14)  The  intersection  of  two  spheres  is  a  circle,  the  plane  of  which  is 

perpendicular  to  the  straight  line  joining  the  centres  of  the 
spheres,  and  the  centre  of  which  is  on  that  line.     §  604. 

6.    Theorems  relating  to  Tangents  to  a  Sphere. 

(1)  A  plane  which  is  perpendicular  to  a  radius  of  a  sphere  at  its 

extremity  is  tangent  to  the  sphere,  and  conversely.     §  601. 

(2)  All  lines  tangent  to  a  sphere  at  one  point  lie  in  a  plane  tangent 

to  the  sphere  at  that  point.     §  602. 
(8)  The  plane  of  two  straight  lines  tangent  to  a  sphere  at  the  same 
point  is  also  tangent  to  the  sphere  at  that  point.     §  603. 

6.    Theorems   relating   to   the    Properties   of    Spherical   Angles, 
Triangles,  and   Polygons. 

(1)  The  angle  formed  by  two  intersecting  arcs  of  great  circles  has 

the  same  measure  as  the  dihedral  angle  formed  by  the  planes 
of  the  circles.    §  613. 

(2)  The  angle  formed  by  two  intersecting  arcs  of  great  circles  has  the 

same  measure  as  the  arc  of  the  great  circle  of  which  the  vertex 
of  the  angle  is  the  pole,  intercepted  between  the  given  arcs 
(produced  if  necessary).     §  615. 

(3)  All  arcs  of  great  circles  drawn  through  the  pole  of  a  given  great 

circle  are  perpendicular  to  the  given  great  circle.     §  616. 

(4)  No  side  of  a  convex  spherical  polygon  can  be  greater  than  a  semi- 

circle.   §  618. 
(6)  The  sum  of  any  two  sides  of  a  spherical  triangle  is  greater  than 

the  third  side.     §  621. 
(6)  The  sum  of  the  sides  of  a  convex  spherical  polygon  is  less  than  a 

great  circle.    §  622. 


Summary]  THE  SPHERE  413 

(7)  An  isosceles  spherical  triangle  and  its  symmetrical  spherical  tri- 

angle are  identically  equal.     §  626. 

(8)  If  two  sides  of  a  spherical  triangle  are  equal,  the  angles  opposite 

those  sides  are  equal.     §  627. 

(9)  If  the  first  of  two  spherical  triangles  is  the  polar  of  the  second, 

then  the  second  is  also  the  polar  of  the  first.     §  631. 

(10)  In  two  polar  triangles,  the  measure  of  any  angle  in  one  is  equal 

to  the  measure  of  the  supplement  of  that  side  in  the  other  of 
which  its  vertex  is  the  pole.     §  633. 

(11)  If  ^  is  any  angle  of  a  spherical  triangle,  and  a'  the  corresponding 

side  of  the  polar  triangle,  then  the  measure  of  a'  is  equal  to  the 
measure  of  the  supplement  of  A.     §  634. 

(12)  If  any  spherical  triangle  is  equiangular,  its  polar  triangle  is  equi- 

lateral, and  conversely.     §  635. 

(13)  If  two  spherical  triangles  on  the  same  sphere,  or  on  equal  spheres, 

are  mutually  equiangular,  their  polar  triangles  are  mutually 
equilateral,  and  conversely.     §  636. 

(14)  The  sum  of  the  angles  of  any  spherical  triangle  is  greater  than 

two  right  angles  and  less  than  six  right  angles.     §  637. 

(15)  A  spherical  triangle  may  have  one,  two,  or  three  right  angles,  or 

one,  two,  or  three  obtuse  angles.    §  639. 

7.  Theorems  on  the  Equality  op  Spherical  Triangles. 

Two  spherical  triangles  lying  on  the  same  or  equal  spheres  are  iden- 
tically equal,  or  symmetrical,  if  they  have  — 

(1)  Three  sides  of  the  one  equal,  respectively,  to  the  three  sides  of 

the  other.     §  623. 

(2)  Two  sides  and  the  included  angle  of  the  one  equal,  respectively, 

to  two  sides  and  the  included  angle  of  the  other.     §  628. 

(3)  One  side  and  the  two  adjacent  angles  of  the  one  equal,  respect- 

ively, to  one  side  and  the  two  adjacent  angles  of  the  other. 
§629. 

(4)  Three  angles  of  the  one  equal,  respectively,  to  the  three  angles 

of  the  other.     §  638. 

8.  Theorems  relating  to  Areas. 

(1)  Two  symmetrical  spherical  triangles  have  equal  areas.    §  646. 

(2)  Any  two  lunes  on  the  same  sphere  or  on  equal  spheres,  having 

equal  angles,  are  equal.     §  648. 

(3)  The  ratio  of  two  lunes  on  the  same  sphere  or  on  equal  spheres  is 

equal  to  the  ratio  of  their  angles.     §  649. 


414  ELEMENTARY  GEOMETRY  [Chap.  IX 

(4)  The  area  of  a  lune  is  to  the  area  of  the  whole  sphere  in  the  same 

ratio  as  the  angle  of  the  lune  is  to  four  right  angles.     §  650. 

(5)  If  two  great  circles  intersect  on  a  hemisphere,  the  two  triangles 

formed  by  their  arcs  and  arcs  of  the  great  circle  bounding  the 
hemisphere  are  together  equal  to  a  lune  having  the  same  angle 
as  the  angle  between  the  great  circles.     §  651. 

(6)  The  area  of  any  spherical  triangle  expressed  in  spherical  units  is 

equal  to  the  spherical  excess  of  the  triangle.  Area  =  A-\-  B  -{■ 
C-180.     §653. 

(7)  It  S  is  the  sum  of  the  angles  of  a  spherical  polygon  of  n  sides,  the 

area  of  the  polygon  is  [S  -  (n  —  2)  180]  spherical  units.    §  654. 

(8)  The  area  of  the  surface  generated  by  a  line-segment  revolving 

about  an  axis  in  its  plane  is  equal  to  the  length  of  the  projec- 
tion of  the  line-segment  on  the  axis  multiplied  by  the  circum- 
ference of  the  circle  whose  radius  is  equal  to  that  segment  of 
the  perpendicular  bisector  of  the  revolving  line-segment  which 
is  intercepted  between  it  and  the  axis.     §  655. 

(9)  The  area  of  a  zone  is  equal  to  the  product  of  its  altitude  and  the 

circumference  of  a  great  circle.     A  =  2  wrh.     §  657. 

(10)  The  area  of  a  zone  is  equal  to  that  portion  of  the  lateral  area  of 

the  enveloping  cylinder  which  is  intercepted  between  the  same 
planes  as  the  zone.     ^  660. 

(11)  The  area  of  a  sphere  is  equal  to  the  product  of  a  diameter  and 

the  circumference  of  a  great  circle.    A  =  2  irrd  =  4  irr^.    §  661. 

(12)  The  areas  of  two  spheres  are  in  the  same  ratio  as  the  squares  of 

their  radii,  or  as  the  squares  of  their  diameters.  A:A'  =  r^:  r'^, 
or  (?2 :  d'i.     §  663. 

(13)  The  area  of  a  sphere  equals  four  times  the  area  of  one  of  its 

great  circles.     §  664. 

(14)  The  area  of  a  sphere  equals  the  lateral  area  of  its  enveloping 

cylinder.    §  665. 

(15)  The  area  of  a  spherical  unit  expressed  in  plane  units  equals 

-^.     §666. 

180      ^ 

9.   Theorems  relating  to  Volumes. 

(1)  The  volume  of  a  sphere  is  equal  to  one-third  the  product  of  its 

area  and  its  radius.     F  =  |  irr^.    §  667. 

(2)  The  volumes  of  two  spheres  are  in  the  same  ratio  as  the  cubes  of 

their  radii,  or  as  the  cubes  of  their  diameters.  V:V'  =  r'^ :  r'^, 
or  d^ :  d'^    §  669. 


Summary]  THE  SPHERE  415 

(3)  The  volume  of  a  spherical  pyramid  or  a  spherical  sector  is  equal 

to  one-third  the  product  of  the  area  of  its  base  and  the  radius 
of  the  sphere.     V=\  Ar.    §  673. 

(4)  The  volume  of  a  spherical  segment  is  equal  to  \h  (jcfi  +  irh'^) 

+  \  irh^,  where  a  and  b  are  the  radii  of  its  bases  and  h  its 
altitude.     §  674. 

10.    Miscellaneous  Theorems. 

(1)  Through  any  four  points  not  lying  in  the  same  plane,  one  and 

only  one  sphere  can  be  passed.     §  605. 

(2)  The  perpendiculars  to  the  four  faces  of  a  tetrahedron,  erected  at 

the  centres  of  the  circumscribed  circles,  all  pass  through  one 
point.     §  608. 

(3)  The  six  planes  perpendicular  to  the  edges  of  a  tetrahedron  at 

their  mid-points  have  one  point  in  common.     §  609. 

(4)  One  and  only  one  sphere  can  be  inscribed  in  any  given  tetra- 

hedron.    §  610. 

(5)  The  planes  which  bisect  the  six  dihedral  angles  of  any  tetrahedron 

have  one  point  in  common.    §  611. 

(6)  The  shortest  line  that  can  be  drawn  on  a  sphere  between  two 

points  is  the  arc  of  a  great  circle,  not  greater  than  a  semicircle, 
joining  the  two  points.     §  642. 


APPENDIX 


o^^c 


The  following  brief  introduction  to  Trigonometry  is  designed 
to  give  to  the  high  school  or  academy  pupil  as  much  of  that 
subject  as  he  may  need  for  a  course  in  Physics  or  Elementary 
Mechanics.  It  contains  no  solid  geometry,  and  may  be  read  as 
soon  as  the  pupil  has  completed  Chapter  V  of  this  text,  or 
earlier  if  desired. 

1.   Trigonometric  Ratios 

Let  BAC  be  any  angle  (for  conven- 
ience at  present,  an  acute  angle), 
and  from  a  point  P  in  one  boundary 
let  a  perpendicular  PM  be  drawn  to 
the  other. 

Then  considering  the  lengths  of  the 
line-segments  AP,  AM,  MP, 

MP 

the  ratio  — —  is  called  the  sine  of  the  angle  A, 

AM 

the  ratio  — —  is  called  the  cosine  of  the  angle  A, 

MP 

the  ratio  — —  is  called  the  tangent  of  the  angle  A. 
AM 

If  the  point  P  were  chosen  differently  on  the  boundary  AC,  and  the 
perpendicular  were  drawn  to  AB^  would  the  ratios  of  the  sides  of  this  new 
triangle  be  equal  to  the  corresponding  ratios  of  the  sides  of  the  triangle 
PAM? 

Or,  if  the  point  P  were  chosen  in  the  boundary  AB  and  the  perpen- 
dicular drawn  to  AC,  would  the  ratios  of  the  sides  be  altered  ? 

416 


APPENDIX  417 

If  the  size  of  the  angle  A  were  changed,  would  the  values  of  these  ratios 
be  changed  ? 

The  answers  to  these  questions  will  show  that  the  values  of  the  sine, 
cosine,  and  tangent  of  the  angle  A  do  not  depend  upon  the  position  of  the 
point  P,  but  do  depend  upon  the  size  of  the  angle. 

From  these  considerations  it  follows  that  equal  angles  have  equal  sines, 
cosines,  and  tangents  ;  unequal  angles  have  unequal  sines,  cosines,  and 
tangents. 

In  the  right  triangle  PAM,  the  side  AP  is  the  hypotenuse ; 
MP,  the  side  opposite  the  vertex  A,  may  be  called  the  per- 
pendicular ;  and  AM,  the  side  adjacent  to  A,  may  be  called  the 
base. 

Then  sine  oiZA  =  ^=  ""'^^  "PP"^'*^^  or  P^'^P'""^'""^'^'^, 
AP       hypotenuse  hypotenuse 

cosine  of  Z  ^  =  ^=  ^^^^  ^'^■1'^^°*  or         ^^^      , 
AP        hypotenuse  hypotenuse 

tangent  of  Z  ^  =  ^=  side  opposite  ^^  perpendicular^ 
AM     side  adjacent  base 

The  reciprocals  of  these  ratios  have  also  received  particular 
names  as  follows : 

=  cosecant  of  Z  ^  =  ^=  hyPotenuse  ^ 
MP     side  opposite 

=    secant  of  Z^    ^  ^^  hypotenuse  ^ 

AM     side  adjacent 

J.    xt  /  A      AM     side  adiacent 

.     o   y   A=  cotangent  of  Z  ^  =  — —  =  -^^ ^ — -. — 

tangent  of  Z  ^  MP     side  opposite 

For  brevity,  we  write  ^sin  A^  instead  of  'sine  of  Z  ^'  but 
this  symbol  should  always  be  understood  to  mean  'the  sine 
of  the  angle  A.' 

Similarly  we  write  cos  A,  tan  A,  cot  A,  sec  A,  cosec  A,  instead 
of  '  cosine  of  Z  A,^  '  tangent  of  Z  A,^  etc. 

It  should  be  noticed  carefully  that  sin  A,  cos  A,  etc.,  are 
mere  numbers,  being  ratios  between  the  lengths  of  certain  line- 
segments. 

2e 


sine  of  Z  ^ 

1 

cosine  of  Z  ^ 

1 

418  ELEMENTARY  GEOMETRY 

To  find  the  trigonometric  ratios  oi  Z  A  (Fig.  1),  PM  was 
drawn  from  a  point  in  one  boundary  perpendicular  to  the  other. 
If  we  wish  to  find  the  ratios  of  Z  P,  we  can  look  upon  A3f 
as  drawn  from  a  point  A  in  one  boundary  of  P,  perpendicular 
to  the  other  boundary,  so  that  A  PAM,  which  is  right-angled 
at  M,  will  give  us  the  ratios  of  both  Z  A  and  Z  P. 

Thus  sin  ^  =  i^  =  cos  P, 

AP 

cos  A  =  — — ■  =  sm  P, 
AP 

.      MP         ^   D 

tan  A  = =  cot  P. 

AM 

That  is  to  say, 

the  sine  of  an  angle  =  the  cosine  of  its  complement, 
the  cosine  of  an  angle  =  the  sine  of  its  complement, 
the  tangent  of  an  arigle  —  the  cotangent  of  its  complement. 

2.     Relations  Among  the  Ratios 

In  the  right  triangle  ABC,  C  being  the  right 
angle, 

W  4-  A(f  =  Aff, 

no  matter  how  large  or  how  small  the  angles 
A  and  ^  may  be. 

Divide  this  relation  through  hy  AB   and  you  obtain 

BO^      ^  ^  A& 

ab'    ab'    aW' 

(S)"*(S)='- 

AT  BC       •      .    AC  . 

Now  —  =  sm  A,  =  cos  A. 

AB  '  AB 

Hence  (sin  Ay  +  (cos  Af  =  1. 


APPENDIX  419 

This  relation  is  usually  written 

sin^  A  -f  COS"  A  =  l, 

and  should   be  read  "  sine-squared  A  plus  cosine-squared  A 
equals  one.'^ 

Also,  in  the  right  triangle  ABO, 

smA^BG_^Aq^BC^^^^^^ 


cos  A     AB     AB     AC 

sin  A 
cos  A 


^.e.  ?i5-4  =  tan  A 


EXERCISES 

1.  Is  sin  A  greater  or  less  than  unity  ?    cos  ^  ?    sec  A  ?    cosec  A  ? 

2.  Suppose  you  keep  the  point  P  fixed  on  the  line  AC  (Fig.  1),  while 
the  line  rotates  so  as  to  make  the  angle  A  increase.  Will  sin  A  increase 
or  decrease  ?    cos  A  ?    tan  A  ? 

3.  Can  tan  A  ever  be  greater  than  unity  ?  Can  it  be  less  than  unity  ? 
For  what  value  of  A  will  it  be  equal  to  unity  ? 

4.  ABC  is  a  triangle  right-angled  at  C,  A  being  the  least  angle.  If 
the  lengths  of  the  sides  are  3  ft.,  4  ft.,  and  5  ft.  respectively,  find  the 
values  of  sin  A^  cos  A^  tan  A^  cos  B,  tan  P,  sin  B. 

5.  Construct  an  angle  whose  sine  is  \. 

Suggestion.    Construct  a  right  triangle  whose  hypotenuse  is  equal  to 
twice  one  side. 

6.  Construct  an  angle  whose  cosine  is  |. 

7.  Construct  an  angle  whose  tangent  is  ^\. 

8.  If  the  sine  of  an  angle  is  y^^,  find  it  cosine  and  its  tangent. 

9.  If  the  cosine  of  an  angle  is  f,  find  its  sine  and  its  tangent. 

10.  If  the  sine  of  an  angle  is  \\^  find  its  cosine  and  its  tangent. 

11.  Divide  the  relation  BC^  +  AC^  =  AB^  through  by  uiC^,  and  what 
formula  do  you  obtain  ? 

12.  Prove  that  (sin  yl  +  cos  ^)2  =  1  +  2  sin  ^  cos  A. 

13.  Prove  that  cos2  A  -  sin2  A  =  2  cos2  A  -  1. 


420 


ELEMENTARY  GEOMETRY 


3.     Numerical  Values  op  the  Ratios   of  Given  Angles 

First,  let  the  right  triangle  ABC  be  such  that  the  side  AC 
equals  the  side  BC. 

Then  Z.A  =  ZB,  and  each  is  half  a  right  angle,  or  45°. 

If 

then 
and 


Hence 


the  measure  of  the  side  AC  =  1, 
the  measure  of  the  side  BC  =  1, 
the  measure  of  the  side  AB=\^2. 
1 


sin  45°=-^=    .7071, 

V2 


cos  45° 


tan  45°  = 


V2 
1 


=    .7071, 


1.0000. 


Why? 


The  decimal  values  given  are  computed  correctly  to  four 
places. 

Next,  suppose  that  AABD  is  equi- 
lateral. Then  Z  ^  =  Z  5  =  Z  D,  and  each 
is  one-third  of  two  right  angles,  or  60°. 

From  B  draw  a  perpendicular  BC  to 
AD.  This  bisects  both  Z  B  and  the  side 
AD. 

The  measure  of  Z  ABC  is  thus  30°. 

If  the  measure  of  each  side  of  the  equilateral   triangle  is 
two,  then  in  A  ABC, 

the  measure  of  the  side  AB  =  2,  of  AC=  1,  and  of  BC  =  V3. 

BG^V^ 

AB       2 

1 


Hence 


sin  60° 


cos  60°  =  —  = 

AB       2 


=    .8660; 
.5000; 


tan  60°  =  ^  =  ^  =  1.7321. 
AC      1 


APPENDIX  421 


Also,  sin  30°  =4?=  \   =    .5000 


cos  30°  =  ^=^=    .8660; 
AB       2 


tan  30°=:^=—=    .5774. 


BO     V3 

By  simple  geometrical  means,  we  have  thus  computed  the 
sine,  cosine,  and  tangent  of  angles  whose  measures  are  30°, 
45°,  and  60°.  It  will  be  very  easy  to  memorize  the  fractional 
values  of  these  ratios  if  the  diagrams  from  which  they  have 
been  derived  are  firmly  fixed  in  the  mind. 

For  example,  it  should  be  noticed  that  in  the  triangle  whose 
angles  are  45°,  45°,  and  90°,  the  opposite  sides  are  proportional 
to  1,  1,  and  V2;  and  in  the  triangle  whose  angles  are  30°,  60°, 
and  90°,  the  opposite  sides  are  proportional  to  1,  V3,  and  2. 

Suppose  that  in  the  right  triangle  CAB,  the  hypotenuse  AB 
remains  of  fixed  length,  while  ZA  is  gradually  decreased  by 
the  rotation  of  AB  about  the  point  A.  The  perpendicular  BC 
will  become  less  and  less,  while  the  base  AC  will  increase  in 
length.     Finally  when  ZA=  0°,  BC=  0,  and  AC  =  AB. 

Hence     sin  0°  =  ^  =  0,  cos  0°  =  ^=  1,  tan  0°  =  —  =  0. 
AB        '  AB  AC 

Similarly  by  rotating  AB  the  other  way  until  ZA  =  90°,  it 
may  be  shown  that  sine  90°  =  1,  cos  90°  =  0,  tan  90°  =  oo. 

The  trigonometric  ratios  of  angles,  or  as  they  are  frequently 
called,  the  trigonometric  functions,  have  been  carefully  com- 
puted, by  one  means  or  another,  for  very  minute  subdivisions 
of  the  angles,  and  have  been  tabulated  so  as  to  be  ready 
for  use. 

.  In  the  following  table  we  give  the  sines,  cosines,  and  tangents 
of  angles  from  0°  to  90°,  at  intervals  of  one  degree,  calculated 
correctly  to  four  decimal  places. 


422 


ELEMENTARY  GEOMETRY 


TABLE   OF  TRIGONOMETRIC   RATIOS 


Degrees 

Sine 

Cosine 

Tangent 

Degrees 

Sine 

Cosine 

Tangent 

0 

.0000 

1.0000 

.0000 

46 

.7193 

.6947 

1.0355 

1 

.0175 

.9998 

.0175 

47 

.7314 

.6820 

1.0724 

2 

.0349 

.9994 

.0349 

48 

.7431 

.6691 

1.1106 

3 

.0523 

.9986 

.0524 

49 

.7547 

.6561 

1.1504 

4 

.0698 

.9976 

.0699 

50 

.7660 

.6428 

1.1918 

5 

.0872 

.9962 

.0875 

51 

.7771 

.6293 

1.2349 

6 

.1045 

.9945 

.1051 

52 

.7880 

.6157 

1.2799 

7 

.1219 

.9925 

.1228 

53 

.7986 

.6018 

1.3270 

8 

•  1392 

.9903 

.1405 

54 

.8090 

.5878 

1.3764 

9 

.1564 

.9877 

.1584 

55 

.8192 

.5736 

1.4281 

10 

.1736 

.9848 

.1763 

56 

.8290 

.5592 

1.4826 

11 

.1908 

.9816 

.1944 

57 

.8387 

.5446 

1.5399 

12 

.2079 

.9781 

.2126 

58 

.8480 

.5299 

1.6003 

13 

.2250 

.9744 

.2309 

59 

.8572 

.5150 

1.6643 

14 

.2419 

.9703 

.2493 

60 

.8660 

.5000 

1.7321 

15 

.2588 

.9659 

.2679 

61 

.8746 

.4848 

1.8040 

16 

•2756 

.9613 

.2867 

62 

.8829 

.4695 

1.8807 

17 

.2924 

.9563 

.3057 

63 

.8910 

.4540 

1.9626 

18 

.3090 

.9511 

.3249 

64 

.8988 

.4384 

2.0503 

19 

.3256 

.9455 

.3443 

65 

.9063 

.4226 

2.1445 

20 

.3420 

.9397 

.3640 

66 

.9135 

.4067 

2.2460 

21 

.3584 

.9336 

.38.39 

67 

.0205 

.3907 

2.3559 

22 

.3746 

.9272 

.4040 

68 

.9272 

.3746 

2.4751 

23 

.3907 

.9205 

.4245 

69 

.9336 

.3584 

2.6051 

24 

.4067 

.9135 

.4452 

70 

.9397 

.3420 

2.7475 

25 

.4226 

.9063 

.4663 

71 

.9455 

.3256 

2.9042 

26 

.4384 

.8988 

.4877 

72 

.9511 

.3090 

3.0777 

27 

.4540 

.8910 

.5095 

73 

.9563 

.2924 

3.2709 

28 

.4695 

.8829 

.5317 

74 

.9613 

.2756 

3.4874 

29 

.4848 

.8746 

.5543 

75 

.9659 

.2588 

3.7321 

30 

.5000 

.8660 

.5774 

76 

.9703 

.2419 

4.0108 

31 

.5160 

.8572 

.6009 

77 

.9744 

.2250 

4.3315 

32 

.5299 

.8480 

.6249 

78 

.9781 

.2079 

4.7046 

33 

.5446 

.8387 

.6494 

79 

.9816 

.1908 

5.1446 

34 

.5592 

.8290 

.6745 

80 

.9848 

.1736 

5.6713 

35 

.5736 

.8192 

.7002 

81 

.9877 

.1564 

6.3138 

36 

.5878 

.8090 

.7265 

82 

.9903 

.1392 

7.1154 

37 

.6018 

.7986 

.7536 

83 

.9925 

.1219 

8.1443 

38 

.6157 

.7880 

.7813 

84 

.9945 

.1045 

9.5144 

39 

.6293 

.7771 

.8098 

85 

.9962 

.0872 

11.4301 

40 

.6428 

.7660 

.8391 

86 

.9976 

.0698 

14.3007 

41 

.6561 

.7547 

.8693 

87 

.9986 

.0523 

19.0811 

42 

.6691 

.7431 

.9004 

88 

.9994 

.0349 

28.6368 

43 

.6820 

.7314 

.9325 

89 

.9998 

.0175 

57.2900  . 

44 

.6947 

.7193 

.9657 

90 

1.0000 

.0000 

QO 

45 

.7071 

.7071 

1.0000 

APPENDIX 


423 


EXERCISES 

1.   In  a  right  triangle  the  hypotenuse  is  25  feet,  and  one  adjacent 
angle  is  32°.     Find  the  other  parts  of  the  triangle. 


Solution.     In  the  right  triangle  ABC^  suppose  Z^  =  32°  and  c  =  25 
feet. 

Then,  to  find  the  side  a, 

-  =  sin  ^  =  sin  32°, 
c 

or  a  =  c .  sin  32°  =  25  x  .5299  =  13.25  feet  nearly. 

To  find  the  side  6, 

-  =  cos  ^  =  cos  32°, 
c 

or  6  =  c .  cos  32°  =  25  x  .8480  =  21.20  feet. 

To  find  the  angle  B, 

5  =  90°  -  ^  =  90°  -  32°  =  58°. 

Test  the  approximate  accuracy  of  these  results  by  showing 


(1) 
(2) 


a2  +  h\ 


tan^  = 


2.  In  a  right  triangle  one  side  is  36  feet  and  the  adjacent  angle  is  54°. 
Find  the  other  parts  of  the  triangle. 

3.  In  an  isosceles  triangle  the  base  is  18  feet  and  the  vertical  angle  48°. 
Find  its  sides. 

4.  A  regular  pentagon  inscribed  in  a  circle  has  a  side  of  20  inches. 
Find  the  radius  of  the  circle. 

5.  Find  the  length  of  the  side  of  a  regular  octagon  circumscribed  about 
a  circle  of  radius  15  inches. 


424  ELEMENT ABY  GEOMETRY 

4.   Ratios  of  Twice  an  Angle  and  Half  an  Angle 

The  following  useful  formulas  can  be  easily  deduced  from 
well-known  geometrical  relations: 

(1)  sin  2  ^  =  2  sin  A  cos  A, 

(2)  cos  2  ^  =  cos2  A  -  sin2  A. 


Let  0  be  the  centre  of  a  circle  of  which  ^(7  is  a  diameter, 
and  ABC  an  inscribed  triangle. 

Z  ABC  is  a  right  angle.  Why  ?  Draw  BD  perpendicular 
to  AC. 

If  Z  CAB  =  A,  Z.  COB  =2  A.  Why  ? 

T?'    ^  '    o  A      ^D      2BD 

First,  ^-2^=05^^10"' 

^J5 '  AC' 

=  2  sin  ^  •  cos  A. 

^^    ^  ^  .      CD     2  CD     AD -DC 

Next,  cos2^  =  — =  ^^  =  -^^— , 

(since  AD  =  radius  +  OD,  and  DC  =  radius  -  OD), 

^AD     DC 
AC     AC' 

^AD   AB_DG   CB 
AB  '  AC     CB  '  AG 

>T  AD     AB   ^    .  DC     CB  . ..   onn\ 

^^^  AB  =  AC^^''^CB==AC'        ^^^'-''^-^ 


APPENDIX  425 

Therefore  -  2  ^  =  (f  J  -  (gj 

=  cos^  A  —  sin^  A. 

If  we  remember  that  sin^^  +  cos^^  =  1,  and  in  the  last 

formula  substitute   for   cos^^l  its   value,  viz.,  1  —  sin^^,  we 

obtain 

cos  2^  =  1-2  sin2  A. 

Or,  if  we  substitute  for  sin^  A  its  value,  1  —  cos^^,  we  obtain 

cos  2  A  =  2  cos^  A-1, 

If  in  these  formulas  P  is  written  in  place  of  2  A,  and  conse- 
quently  —  in  place  of  A,  they  become 

P       P 

sin  P  =  2  sin  —  cos  — - 

cos  P  —  cos^—  —  sin^— » 

Z  LI 

=  l-2sin2|, 
=  2  008^:^-1. 

LI 

The  latter  two  formulas  may  be  rewritten  in  the  form, 
2  siii2^=l-cos  J*. 

Li 

2  cos2^=l  +  cosI*. 

LI 

EXERCISES 

sin  lA 


1.   From  the  last  diagram  produce  the  formula  tan  A 


1  +  cos  2^ 

2.  Deduce  the  values  of  the  sine,  cosine,  and  tangent  of  16°  from  the 
formulas  given  above,  and  compare  your  results  with  those  given  in  the 
table. 

3.  Find  in  a  similar  way  the  values  of  the  sine,  cosine,  and  tangent 
of  22^  degrees. 


426  ELEMENTARY  GEOMETRY 

5.   Ratios  of  an  Obtuse  Angle 

Suppose  Z.BAG  is   obtuse.     Then   the   perpendicular   PM 
drawn  from  a  point  in  one        q 
boundary    meets,    not    the 
other    boundary,    but    the 
boundary    produced    back- 
ward through  the  vertex. 

In  that  case,  the  base  AM,  _ 
i.e.  the  intercept  between 
the  vertex  and  the  foot  of  the  perpendicular,  is  considered  to  be 
negative.  The  perpendicular  PM  and  the  hypotenuse  AP  are 
considered  positive,  just  as  in  the  case  of  an  acute  angle. 

The  definitions  of  the  functions  of  the  obtuse  angle  BAG 
are  the  same  as  for  the  acute  angle  BAC. 

Hence,  sine  of  the  obtuse  angle  BAC  = ,  and  is  positive. 

Cosine  of  the  obtuse  angle  BAC  =  -——,  and  is  negative,  since 

AM'\^  negative. 

MP 

Tangent  of  the  obtuse  angle  BAC  — ,  and  is   negative, 

AM 
Since  AM  is  negative. 

The  functions  of  an  acute  angle  are  all  positive  since  the 
perpendicular  PM  falls  upon  the  other  boundary,  and  not 
upon  that  boundary  produced  backward. 

The  thing  to  be  remembered  in  this  connection  is  that  if  the 
perpendicular  drawn  from  a  point  in  one  boundary  of  the  angle 
meets  the  other  boundary,  the  base  is  considered  positive  ;  but  if  it 
meets  the  other  boundary  produced  backward  through  the  vertex, 
the  base  is  considered  7iegative. 

.                                                                MP 
The  sine  of  the  obtuse  angle  BAC  is  ;  but  this  ratio  is 

also  the  sine  of  the  supplementary  angle  B'AC,  and  for  both 
angles  the  ratio  is  positive. 

Hence,  the  sine  of  any  angle  equals  the  sine  of  its  supplement. 

The  cosine  of  the  obtuse  an^rle  BAC  is  ,  which  is  also 

AP 


APPENDIX  421 

the  cosine  of  the  supplementary  angle  B'AC.  But  for  the 
latter  angle  the  base  AM  must  be  considered  positive,  since 
the  perpendicular  PM  meets  the  boundary  AB'  of  that  angle, 
while  for  the  former  angle  AM  is  negative. 

That  is,  the  cosine  of  an  obtuse  angle  is  equal  in  magnitude 
to  the  cosine  of  its  supplement,  but  is  negative.  In  other 
words,  the  cosine  of  any  angle  equals  minus  the  cosine  of  its 
supplement. 

Similarly,  the  tangent  of  an  obtuse  angle  is  equal  in  magni- 
tude to  the  tangent  of  its  supplement,  but  is  negative.  In 
other  words,  the  tangent  of  any  angle  equals  minus  the  tangent 
of  its  supplement. 

EXERCISES 

1.  Write  the  values  of  sin  120°,  cos  135°,  tan  150°. 

2.  What  are  the  values  of  the  sine,  cosine,  and  tangent  of  an  angle 
of  105°  ? 

3.  What  is  the  difference  in  value  between  sin  45°  and  sin  135° ; 
between  cos  45^  and  cos  135°  ? 

4.  Find  the  value  of  sin  2  A  when  cos  A  =  ^. 

5.  Find  the  value  of  cos  2  A  when  sin  A  =  j%. 

6.  If  tan  A  =  j^,  find  the  value  of  sin  A,  cos  A,  sin  2  A,  cos2  ^. 

7.  If  a  chord  8  ft.  in  length  is  placed  in  a  circle  of  5  ft.  radius,  find 
approximately  the  size  of  the  angle  it  subtends  at  the  centre,  and  the 
length  of  its  arc. 

8.  The  sides  of  a  right  triangle  are  5,  12,  and  13  feet,  respectively. 
Find  the  angles  of  the  triangle.  The  mid-point  of  the  side  12  is  joined  to 
the  opposite  vertex ;  into  what  two  parts  does  the  line  so  drawn  divide 
the  opposite  angle  ? 

9.  The  upper  part  of  a  tree  broken  off  with  the  wind  makes  an  angle 
of  30°  with  the  ground,  and  the  distance  from  the  root  to  the  point  where 
the  top  of  the  tree  touches  the  ground  is  50  ft.  What  was  the  height  of 
the  tree  ? 

10.  The  angular  elevation  of  the  top  of  a  chimney  when  viewed  from 
one  position  is  32°,  and  on  walking  96  ft.  in  a  straight  lijae  toward  the  foot 
of  the  chimney  the  angular  elevation  of  the  top  becomes  48°.  Find  the 
height  of  ih^  chimnejr, 


428 


ELEMENTARY  GEOMETRY 


6.    Relations  between  the  Sides  and  Angles  of  a 
Triangle 


1.    In  any  triangle 


sin  A      sin  B      sin  C 


Then 
Also 


=  sin  JB ;  hence  AD  =  AB  sinB  =  c  sin  B. 
=  sin  C;  hence  AD  =  AC  sinC  =b  sin  C. 


D       «  G 

From  any  vertex  ^  of  A  ABC  draw  the  perpendicular  AD 
to  the  opposite  side. 
AD 
AB 
AD 
AC 
Therefore  c  sin  B  =  b  sin  C. 

Dividing  through  by  be  gives 

sin  B  _  sin  C 
b     ~     c     ' 
If  the  perpendicular  were  drawn  from  the  vertex  B  to  the 
opposite  side,  we  should  find  in  just  the  same  way  that 
sin  A      sin  C 


Therefore 


a  c 

sin  A     sin  B     sin  C 


EXERCISES 

1.  What  relation  would  be  produced  by  drawing  the  perpendicular 
from  the  vertex  C  ? 

2.  In  a  triangle  a  =  18  ft.,  B  =  27°,  C  =  65°.      Find  the  other  parts  of 
the  triangle. 

3.  If  A  =  86°,  a  =  15  ft.  and  b  =  24,  show  that  B  might  have  either  of 
two  values,  and  find  those  values  approximately. 


APPENDIX 
2.    In  any  triangle  d^  =  a^  ^-b^ -2ah  cos  C 


429 


First,  when  the  angle  C  is  acute, 

Al^  =  BC''  +  10^-2  BC  •  DC. 
But  DO  =  AC  cos  C. 


(Art.  321.) 


Therefore      A^  =  B^  +  AC""  -  2  BC  -  AC  cos  O, 
or  <^  =  a^  -{-  b^  —  2ab  cos  (7. 

Next,  when  the  angle  C  is  obtuse, 

AB'^BC'-h  AC' +2  BC  -  CD.  (Art.  322.) 

But  CD  =  AC  cos  ACD  =  -AC  cos  ACB. 

Therefore      AB"  =  BG""  +  JZ''  -  2  50  •  ^(7  cos  O, 
or  c^  =  a^  +  6^  —  2  a6  cos  C 

Similarly,  6^  =  c^  +  a^  —  2  ca  cos  5, 

a2  =  ^,2_|_^2_25ccosA 


These    formulas    may    be     written,    cos    A 


cos  B  = 


C2  -I-  a2  _  1)2 

2ca 


,  cos  0  = 


a^-^b' 


2  6c       ' 


2a6 


430  ELEMENTARY  GEOMETRY 


3.   In  any  triangle  siii-^  =  'y^ y- — —• 


2       ^  be 

2  sin^^  =  1  -  cos  ^  (Page  425.) 

=  1  _  ^'  +  c^  -  g'  ^  2  6c  -  6^  -  c^  +  g^ 
2  be  2  be 

^  g^  -(6  -  c)^  ^  (g  -  6  +  c)(a  +  b  -  c) 
2  be  2  be 

If,  as  in  Art.  324,  we  let  g  -h  6  +  c  =  2  s, 
then  g  -  5  +  c  =^  2  6'  -  2  6  =  2(.9  -  6), 

a-i-b-e=2s-2e  =  2(s  —  c). 

Hence  2  sin^  ^  =  ^(^^I^Uli^, 

2  2&C 


and  sin4  =  J(i^Mi^. 

2       >/  be 


4.  /?i  g^i?/  triangle  cos  -^  =  'y  "^ ^' 


6c 

2  cos-  :^  =  1  +  cos  ^  (Page  425.) 

^^      6^-f  c^-g^^  2  6c +  6- +  c'-a- 
2  6c     ~  2  6c 

^  (6  +  ef  -  g.2  ^  (6  +  c  +  g.)(64-c  -  g) 
2  6c  2  6c 


_2s'2(s-a) 
2  6c 

Hence  cos--^  =  "^'^  -  ^\ 

2       .    6c 

and  cos^-a/'<^-^I 

2  -\      6c 


APPENDIX 


431 


5.   In  any  triangle  a  =  6  cos  C  +  c  cos  B, 


B  D        a  G        B  C 

From  the  vertex  A  draw  the  perpendicular  AD  to  the  oppo- 
site side. 

In  case  B  and  C  are  both  acute  angles, 
a  =  BD-\-DC 
=  AB  cos  B  -f  AC  cos  C 
=  c  cos  B  +  h  cos  C. 
In  case  one  angle,  say  C,  is  obtuse, 
a  =  BD-CD 
=  ^^  cos  5-^0  cos  AGD. 
But    cos  ^  CZ)  =  — cos  ^(7^,  since  these  are  supplementary- 
angles.     The  angle  ACB  is  Z  (7  of  the  given  triangle. 
Therefore  a  =  AB  go^  B-\-  AG  cos  C. 

=  c  cos  B  -\-h  cos  C 


EXERCISES 

1.  In  a  triangle  whose  sides  are  7  ft.,  9  ft.,  and  12  ft.,  find  the  angles 
(1)  from  their  cosines,  (2)  from  the  sine  of  half  of  each  angle. 

2.  Two  adjacent  sides  of  a  triangle  are  11  and  15  ft.,  respectively,  and 
the  included  angle  is  60°.     Find  the  other  parts. 

3.  The  sides  of  a  triangle  are  10  ft.,  13  ft.,  and  15  ft.,  respectively ; 
find  the  greatest  and  the  least  angle. 

4.  Two  angles  of  a  triangle  are  36°  and  63°,  respectively,  and  the 
greatest  side  is  20  ft. ;  find  the  remaining  parts  of  the  triangle. 

5.  Two  sides  of  a  triangle  are  15  ft.  and  18  ft.,  respectively,  and  the 
included  angle  is  105°.    Find  the  third  side. 


432  ELEMENTARr  GEOMETRY 

7.     Areas  of  Triangles 
1.    The  area  of  any  triangle  is  equal  to  i  be  sin  A, 
B 


The  area  of  a  triangle  =  Ihhj  where  h  is  the  side  AC,  and  h 


the  altitude. 
But 


(Art.  306.) 


h 


=  sin  A,  or  h  =  c  sin  A. 


Substituting  this  value  of  h,  we  have 


area  =  ^bc  sin  A. 


Similarly  it  may  be  shown  that 

area  =  ^  ab  sin  (7, 


-^  ac  sin  B. 


2.    The  area  of  any  triangle  equals  y/s{8  —  a){s  —  b)(s  —  c), 
area  =  ^bc  sin  ^, 

=  i  6c  .  2  sin  ^  cos  ^.  (Page  425.) 


But 


sin 


V(s  — 6)(s  — c)        -I         A        \s(s  —  a 


Therefore     area  =  6c J5^ME«).    J?5^, 
^  oc  ^      oc 


=  Vs(s  —  a){s  —  b)(s  —  c). 


APPENDIX 


433 


8.     Inscribed  and  Circumscribed  Circles 

1.    The  radius  of  the  circle  inscribed  in  a  triangle  equals  — > 

s 

where  A  denotes  the  area  of  the  triangle,  and  s,  half  the  sum  of 
the  sides. 


Let  0  be  the  centre  of  the  inscribed  circle,  and  OF,  OG,  OH, 
radii  drawn  to  the  points  of  contact,  and  therefore  perpen- 
dicular to  the  sides  of  the  given  triangle.     Join  OA,  OB,  OC. 

A  OBG  +  A  OCA  +  A  OAB  =  A  ABC  whose  area  is  A. 
Area  of  A  OBC  =  ^  OF-  BC=\r'a. 
Area  of  AOCA  =  ^OG  -  CA=  ^r-h. 
A.xQ2,oiAOAB  =  ^OH'AB  =  \r'C. 
Therefore      A=:\  r(a  +  6  -f  c), 
=  r '  s 

Therefore  r  =  — 

s 

In  a  similar  way  it  may  be  shown  that  for  a  circle  tangent 
to  the  side  a  and  to  the  sides  b  and  c  produced, 

A 


s  —  a 


2f 


434  ELEMENTARY  GEOMETRY 

2.  : 

equals  ^^ 


2.    Tlie   radius  of  the   circle  circumscribed   about  a  triangle 
ahc 


Let  0  be  the  centre  of  the  circumsoribed  circle,  and  OD  be 
drawn  perpendicular  to  the  side  BC. 

Then  Z  BOD  =  i  Z  BOC  =  Z  A  (Art.  178.) 

— —  =  sin  BOD  =  sin  A, 
BO 


or 

^a  =  R  sin  A. 

Therefore 

^  =  o     •         .  =  o 

abc 

abc 

A                A        A 

(Page  432.) 

EXERCISES 

1.  If  the  sides  of  a  triangle  are  56,  65,  and  33  ft.,  find  the  greatest 
angle  and  the  area  of  the  triangle. 

2.  The  sides  of  a  parallelogram  are  15  and  17  ft.,  and  one  of  its 
smaller  angles  is  42°.     Find  its  area  and  the  lengths  of  its  diagonals. 

3.  The  sides  of  a  triangular  field  are  119,  111,  and  92  yds.     Show 
that  its  area  is  10  sq.  yds.  less  than  an  acre. 

4.  Two  angles  of  a  triangle  are  42°  and  68°,  respectively,  and  the  least 
side  is  20  ft.     Find  its  area. 

5.  The  sides  of  a  triangle  are  11  ft.,  13  ft.,  and  18  ft.,  respectively. 
Find  the  radius  of  the  inscribed  circle. 

6.  Two  sides  of  a  triangle  are  21  ft.  and  24  ft.,  respectively,  and  the 
included  angle  is  110°.    Find  its  area. 


MENSURATION   FORMULAS 


In  the  following  table,  unless  otherwise  specified,  A  denotes 
the  area ;  a,  b,  c,  the  sides  of  a  triangle ;  6,  the  base ;  c,  the 
hypotenuse  of  a  right  triangle ;  r,  the  radius  ;  d,  the  diameter ; 
I,  the  lateral  edge ;  s,  the  slant  height ;  C,  the  circumference ; 
P,  the  perimeter ;  L,  the  lateral  area ;   F,  the  volume. 


1. 

In  any  parallelogram 

A=  bh. 

Art.  305 

2. 

In  aright  triangle     c'2  =  ^2  +  &2, 

Art.  317 

3. 

In  any  triangle 

;       A  =  l  bh. 

Art.  306 

A  =Vs(s  ~  a)(s -  b)(s  -  c). 

Art.  324,  Pg.  432 

A  =  Ibc  sin  A. 

Pg.  432 

a2  ^  52  +  c2  -  2  &c  cos  A. 

Pg.  429 

sin  A     shi  B      sin  C 
a             b             c 

Pg.  428 

sin^=V(^-^)(^-^). 
2       >            6c 

Pg.  430 

4. 

In  a  circle 

C  =  Trd  =  2  Trr. 

Art.  362 

A  =  -1  Cr  =  7rr2  =  1 7r^2. 

Art.  365 

5.  In  a  regular  polygon 

A  =  ^  Pa,  where  a  is  the  apothem. 

6.  In  a  parallelepiped  or  any  prism 

L  =  IP, 

where  P  is  the  perimeter  of  a  right  section. 

V=hA, 

where  A  is  the  area  of  the  base. 
435 


Art.  353 


Art.  479 


Art.  507 


436  ELEMENTARY  GEOMETRY 

7.  In  a  regular  pyramid 

where  P  is  the  perimeter  of  the  base.  Art.  512 

where  A  is  the  area  of  the  base.  Art.  523 

8.  In  a  circular  cylinder 

L  =  IP, 

where  P  is  the  perimeter  of  a  right  section.  Art.  560 

V=  irrVi.  Art.  562 

9.  In  a  right  circular  cone 

L  =  irrs, 


where  r  is  the  radius  of  the  base. 

Art.  577 

F  =  1  TrrVi. 

Art.  579 

10.   In  a  sphere                         A  =  4  Tr\ 

Art.  661 

^  =  720  spherical  units. 

Art.  652 

F  =  |7rr3  =  |^r. 

Art.  667 

11.   In  a  spherical  segment 

' 

L  =  2  wrh. 

Art.  657 

F=i/i(7ra2  +  ^62)  +  i, 

^h% 

where  a  and  b  are  the  radii  of  the  two  bases. 

Art.  674 

INDEX   TO   DEFINITIONS 

The  numbers  refer  to  pages 


Acute  angle,  11. 
Adjacent,  angles,  8. 

polygons,  194. 
Altitude,  197,  198,  311,  329. 
Angle,  6. 

acute,  11. 

cosine  of,  416, 

dihedral,  284. 

in  an  arc.  111. 

obtuse,  11. 

polyhedral,  296. 

reflex,  12. 

right,  9. 

sine  of,  416. 

spherical,  380. 

straight,  12. 

tangent  of,  416. 

trihedral,  296. 
Angles,  adjacent,  8. 

alternate,  62. 

complementary,  67. 

supplemetary,  11. 

of  a  triangle,  7. 

vertical,  8. 
Apothem,  239. 
Arc  of  a  circle ,  100. 

length  of,  245. 
Arcs,  similar,  249. 
Area,  195,  245,  394. 
Axiom,  16. 
Axis,  of  a  circle,  372. 


Axis,  radical,  170. 
of  symmetry,  88. 

Centroid  of  a  triangle,  80. 
Chord,  of  a  circle,  100. 

of  contact,  119. 
Circle,  14. 

arc  of,  100. 

centre  of,  14. 

chord  of,  100. 

diameter  of,  14. 

radius  of,  14. 

secant  of,  115. 

sector  of,  101. 

segment  of,  101. 

tangent  of,  116. 
Circles,  coaxial  system  of,  170. 

concentric,  18. 

in  contact,  130. 
Circular,  cone,  361. 

cylinder,  354. 
Circumference  of  a  circle,  244. 
Circumscribed  figure,  1 19. 
Closed  surface,  310. 
Coaxial  circles,  170, 
Commensurable  magnitudes,  141. 
Common  tangents,   direct   and   in- 
verse, 133. 
Concentric,  circles,  18. 

spheres,  370. 
Concurrent  lines,  79. 
437 


438 


INDEX 


Coney clic  points,  114. 
Cone,  361. 

frustum  of,  365. 

right  circular,  362. 
Conical  surface,  361. 
Conjugate  arcs,  101. 
Contact,  chord  of,  119, 

circles  in,  130. 
Continued  proportion,  152. 
Continuity,  principle  of,  125. 
Converse  theorems,  32. 
Corollary,  35. 
Corresponding  angles,  62. 
Cosine  of  an  angle,  416. 
Cube,  317. 
Cylinder,  354. 

right  circular,  354. 
Cylindrical  surface,  353. 

Decagon,  233. 
Diagonal,  71. 
Diagonals,  principal,  252. 
Diameter,  14. 
Dihedral  angle,  284. 

plane  angle  of,  285. 
Distance  from  a  point,  to  a  straight 
line,  53. 

to  a  plane,  275. 
Dodecagon,  233. 
Dodecahedron,  311. 

Equilateral  triangles,  22. 
Extreme  and  mean  ratio,  177. 
Extremes,  149. 

Figure,  circumscribed,  119. 
closed,  13. 
inscribed.  111. 
plane,  5.  * 

rectilinear,  13. 
solid,  263. 


Figures,  identically  equal,  9. 

isoperimetric,  228. 

symmetrical,  88. 
Frustum,  of  a  cone,  365. 

of  a  pyramid,  328. 

Great  circle,  371. 

Harmonic  division,  186. 
Hexagon,  233. 

principal  diagonals  of,  252. 
Hexahedron,  311. 
Homologous  sides,  30. 
Hypotenuse,  44. 

Icosahedron,  311. 

Incommensurable  magnitudes,  141. 
Indirect  proof,  47. 
Inscribed  figure,  111. 
Inverse  points,  88. 
Isoperimetric  figures,  228. 
Isosceles,  triangle,  22. 
trihedral  angle,  302. 

Length  of  an  arc,  245. 

Limit,  147. 

Line,  oblique  to  a  plane,  266. 

parallel  to  a  plane,  266. 

perpendicular  to  a  plane,  266. 

straight,  3. 
Line-segments,  3. 
Lines,  concurrent,  79. 

parallel,  61. 

perpendicular,  9. 

skew  or  gauche,  271. 

Locus,  50. 

Lune,  395. 

angle  of,  395. 
■>  • 

Magnitude,  16. 
Maximum  value,  111,  226. 


INDEX 


439 


Mean  proportional,  152. 

Means,  149. 

Measure,  140. 

Median  of  a  triangle,  79. 

Minimum  value,  111,  226. 

Multiple,  140. 

Octagon,  233. 
Octahedron,  311. 
Orthocentre  of  a  triangle,  83. 

Parallel,  lines,  61. 

planes,  266. 
Parallelepiped,  317. 

rectangular,  317. 

right,  317. 
Parallelogram,  74. 

altitude  of,  197. 
Pentagon,  233. 
Pentahedron,  311. 
Perimeter,  79. 
Perpendicular,  to  a  line,  9. 

to  a  plane,  266. 
Plane  surface,  4. 
Points,  coney clic,  114. 

inverse,  88. 
Polar  spherical  triangles,  387. 
Poles  of  a  circle,  372. 
Polygon,  71,  73. 

convex,  71,  73. 

diagonal  of,  71. 

opposite  sides  of,  72. 

perimeter  of,  79. 

regular,  233. 

spherical,  382. 
Polygons,  adjacent,  194. 

mutually  equiangular,  161. 

similar,  161. 
Polyhedral  angle,  296. 
Polyhedral  angles,  symmetrical,  297. 
Polyhedron,  310. 

circumscribed,  378. 


Polyhedron,  convex,  310. 

inscribed,  378. 

regular,  342. 
Polyhedrons,  similar,  339. 
Postulate,  15. 

Principle  of  continuity,  126. 
Prism,  311. 

truncated,  315. 
Problem,  17. 
Projection,  212,  292. 
Proportion,  149. 
Pyramid,  328. 

Quadrant,  372. 
Quadrilateral,  73. 

Radical  axis,  170. 

Radius,  14. 

Ratio,  142,  144,  148. 

extreme  and  mean,  177. 
Rectangle,  74. 
Regular  polygon,  233. 

apothem  of,  239. 

centre  of,  239. 

radius  of,  239. 
Regular  pyramid,  328. 

slant  height  of,  328. 
Regular  polyhedron,  342. 
Rhombus,  74. 
Right  angle,  9. 
Right  triangle,  44. 
Right  circular,  cone,  362. 

cylinder,  354. 

prism  311. 

Scalene  triangles,  22. 
Secant  of  a  circle,  115. 
Section  of  a  surface,  310. 
Sector,  of  a  circle,  101. 

of  a  sphere,  406. 
Segment,  of  a  circle,  101, 

of  a  sphere,  401. 


440 


INDEX 


Semicircle,  101. 
Similar,  arcs,  249. 

polygons,  161. 

polyhedrons,  339. 

sectors,  250. 

segments,  249. 
Sine  of  an  angle,  416. 
Sphere,  370. 
Spherical,  angle,  380. 

excess,  391. 

polygon,  382. 

pyramid,  406. 

sector,  406. 

segment,  401. 

triangle,  382. 

unit  of  surface,  397. 
Square,  74. 
Surface,  closed,  310. 

conical,  361. 

cylindrical,  353. 

plane,  4. 
Symmetrical,  figures,  88. 

polyhedral  angles,  297. 

spherical  triangles,  384. 
Symmetry,  axis  of,  88. 

Tangent,  of  an  angle,  416. 
to  a  circle,  116. 
to  a  cone,  362. 


Tangent,  to  a  cylinder,  355. 

to  a  sphere,  375. 
Tangents,  common,  direct  and  in- 
verse, 133. 
Tetrahedron,  311. 
Theorem,  17. 
Third  proportional,  152. 
Transversal,  62. 
Trapezium,  74. 
Trapezoid,  74. 

isoceles,  77. 
Triangle,  5. 

altitude  of,  198. 

centroid  of,  80. 

equilateral,  22. 

isosceles,  22. 

median  of,  79. 

right,  44. 

scalene,  22. 

spherical,  382. 
Trihedral  angle,  296. 

isosceles,  302. 
Truncated,  prism,  315. 

pyramid,  328. 

Vertex,  5,  71,  310. 

Volume  of  a  polyhedron,  315,  324. 

Zone,  401. 


■     ■     -..;.. 

7^-' 

UNIVERSITY  OF  CALIFORNIA  LIBRARY 
BERKELEY 

Return  to  desk  from  which  borrowed. 
This  book  is  DUE  on  the  last  date  stamped  below. 

Novsed  1947 
SEP  19  m 

isMay'SSKV 
"''   '1955LU 

26feb'C0CT 

SEP  1  7  1953 

Esxeijsfii 

SEP  10 1953  Ltt 

lO0ct'53BW 

. 

•CT10l95SILt 

LD  21-100tn-9,'47(A5702sl 

3)476 

YB   17294 


^18229 


Hi 


THE  UNIVERSITY  OF  CALIFORNIA  LIBRARY 


:|,iiilijlJiilil!llililiili!li 


